Does this look like a correct Bandpass filter?
up vote
1
down vote
favorite
A project task assigned me to make a simple series RLC bandpass filter with values B = 73kHz & freq center = 4.8MHz known.
I use equations B=R/L and fc=1/(2pisqrt(LC)) to get the RLC values.
Is the graph what I should expect? The center frequency lines up properly.. and I think the bandwidth is okay because it is so small (73kHz).
/edit
Here is a more zoomed-in version of my graph
filter electrical engineering
asked Nov 22 at 17:07
Austin Brown
314
|
show 1 more comment
up vote
1
down vote
favorite
A project task assigned me to make a simple series RLC bandpass filter with values B = 73kHz & freq center = 4.8MHz known.
I use equations B=R/L and fc=1/(2pisqrt(LC)) to get the RLC values.
Is the graph what I should expect? The center frequency lines up properly.. and I think the bandwidth is okay because it is so small (73kHz).
/edit
Here is a more zoomed-in version of my graph
filter electrical engineering
asked Nov 22 at 17:07
Austin Brown
314
Set your x-axis to a log scale and you'll see something more like what you would see in a textbook.
– John D
Nov 22 at 17:20
I would have gone with $C_1=100:text{pF}$ as a starting value and used Q and the center frequency's relationship to LC to work out the values. However, the values you have "feel" they are in the right ballpark. Your Q seems a bit high. But that's easily fixed.
– jonk
Nov 22 at 18:06
To verify if your bandwidth is exactly 73kHz, expand your x scale (don't make it a log scale) and look for the points where the response is 3dB down from the peak. They should be 73kHz apart.
– TimWescott
Nov 22 at 18:28
@TimWescott I tested this and this gives me 12kHz apart only. Why is it that you choose 3dB? :)
– Austin Brown
Nov 22 at 18:36
I agree with @TimWescott about not making it a log scale when making BW measurements, I was just suggesting the log scale because many textbook bandpass curves are plotted on a log scale and the shape looks more familiar that way. (In case that was one of the concerns with the question "Is the graph what I should expect?"
– John D
Nov 22 at 19:00
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A project task assigned me to make a simple series RLC bandpass filter with values B = 73kHz & freq center = 4.8MHz known.
I use equations B=R/L and fc=1/(2pisqrt(LC)) to get the RLC values.
Is the graph what I should expect? The center frequency lines up properly.. and I think the bandwidth is okay because it is so small (73kHz).
/edit
Here is a more zoomed-in version of my graph
filter electrical engineering
asked Nov 22 at 17:07
Austin Brown
314
A project task assigned me to make a simple series RLC bandpass filter with values B = 73kHz & freq center = 4.8MHz known.
I use equations B=R/L and fc=1/(2pisqrt(LC)) to get the RLC values.
Is the graph what I should expect? The center frequency lines up properly.. and I think the bandwidth is okay because it is so small (73kHz).
/edit
Here is a more zoomed-in version of my graph
filter electrical engineering
filter electrical engineering
asked Nov 22 at 17:07
Austin Brown
314
asked Nov 22 at 17:07
Austin Brown
314
edited Nov 22 at 18:44
asked Nov 22 at 17:07
Austin Brown
314
asked Nov 22 at 17:07
Austin Brown
314
asked Nov 22 at 17:07
Austin Brown
314
314
Set your x-axis to a log scale and you'll see something more like what you would see in a textbook.
– John D
Nov 22 at 17:20
I would have gone with $C_1=100:text{pF}$ as a starting value and used Q and the center frequency's relationship to LC to work out the values. However, the values you have "feel" they are in the right ballpark. Your Q seems a bit high. But that's easily fixed.
– jonk
Nov 22 at 18:06
To verify if your bandwidth is exactly 73kHz, expand your x scale (don't make it a log scale) and look for the points where the response is 3dB down from the peak. They should be 73kHz apart.
– TimWescott
Nov 22 at 18:28
@TimWescott I tested this and this gives me 12kHz apart only. Why is it that you choose 3dB? :)
– Austin Brown
Nov 22 at 18:36
I agree with @TimWescott about not making it a log scale when making BW measurements, I was just suggesting the log scale because many textbook bandpass curves are plotted on a log scale and the shape looks more familiar that way. (In case that was one of the concerns with the question "Is the graph what I should expect?"
– John D
Nov 22 at 19:00
|
show 1 more comment
Set your x-axis to a log scale and you'll see something more like what you would see in a textbook.
– John D
Nov 22 at 17:20
I would have gone with $C_1=100:text{pF}$ as a starting value and used Q and the center frequency's relationship to LC to work out the values. However, the values you have "feel" they are in the right ballpark. Your Q seems a bit high. But that's easily fixed.
– jonk
Nov 22 at 18:06
To verify if your bandwidth is exactly 73kHz, expand your x scale (don't make it a log scale) and look for the points where the response is 3dB down from the peak. They should be 73kHz apart.
– TimWescott
Nov 22 at 18:28
@TimWescott I tested this and this gives me 12kHz apart only. Why is it that you choose 3dB? :)
– Austin Brown
Nov 22 at 18:36
I agree with @TimWescott about not making it a log scale when making BW measurements, I was just suggesting the log scale because many textbook bandpass curves are plotted on a log scale and the shape looks more familiar that way. (In case that was one of the concerns with the question "Is the graph what I should expect?"
– John D
Nov 22 at 19:00
Set your x-axis to a log scale and you'll see something more like what you would see in a textbook.
– John D
Nov 22 at 17:20
Set your x-axis to a log scale and you'll see something more like what you would see in a textbook.
– John D
Nov 22 at 17:20
I would have gone with $C_1=100:text{pF}$ as a starting value and used Q and the center frequency's relationship to LC to work out the values. However, the values you have "feel" they are in the right ballpark. Your Q seems a bit high. But that's easily fixed.
– jonk
Nov 22 at 18:06
I would have gone with $C_1=100:text{pF}$ as a starting value and used Q and the center frequency's relationship to LC to work out the values. However, the values you have "feel" they are in the right ballpark. Your Q seems a bit high. But that's easily fixed.
– jonk
Nov 22 at 18:06
To verify if your bandwidth is exactly 73kHz, expand your x scale (don't make it a log scale) and look for the points where the response is 3dB down from the peak. They should be 73kHz apart.
– TimWescott
Nov 22 at 18:28
To verify if your bandwidth is exactly 73kHz, expand your x scale (don't make it a log scale) and look for the points where the response is 3dB down from the peak. They should be 73kHz apart.
– TimWescott
Nov 22 at 18:28
@TimWescott I tested this and this gives me 12kHz apart only. Why is it that you choose 3dB? :)
– Austin Brown
Nov 22 at 18:36
@TimWescott I tested this and this gives me 12kHz apart only. Why is it that you choose 3dB? :)
– Austin Brown
Nov 22 at 18:36
I agree with @TimWescott about not making it a log scale when making BW measurements, I was just suggesting the log scale because many textbook bandpass curves are plotted on a log scale and the shape looks more familiar that way. (In case that was one of the concerns with the question "Is the graph what I should expect?"
– John D
Nov 22 at 19:00
I agree with @TimWescott about not making it a log scale when making BW measurements, I was just suggesting the log scale because many textbook bandpass curves are plotted on a log scale and the shape looks more familiar that way. (In case that was one of the concerns with the question "Is the graph what I should expect?"
– John D
Nov 22 at 19:00
|
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
2
down vote
The ratio of centre frequency to bandwidth is called Q and Q, for a series resonant circuit, is also dictated by R, L and C using this formula: -
$$Q = dfrac{1}{R}sqrt{dfrac{L}{C}}$$
Plugging in your circuit values I get a Q of 413. If I considered the centre-frequency to bandwidth ratio, I get 65.75 so, it looks like you will be "sharper" in resonance than your specification demands and this can cause problems.
I need to know if the method (using the equations I listed) is the correct way to determine my RLC values
The ratio of R to L does give you the bandwidth but that bandwidth is in radians per second; you have assumed it is in hertz hence, the value of Q of 413 (gleaned from my formula and your values) is precisely $2pi$ times higher than 65.7 ($f_0$ to bandwidth ratio) and that is your only mistake. The formula for resonant frequency is correct.
answered Nov 22 at 17:35
Andy aka
238k10174406
1
That's about what I get. I think this is totally a theoretical exercise for the OP and their Q seems excessive, just as you point out. I think this is the right kind of response, though.
– jonk
Nov 22 at 18:07
Yes this exercise is 100% theoretical.
– Austin Brown
Nov 22 at 18:37
I need to know if the method (using the equations I listed) is the correct way to determine my RLC values, and also that my graph is okay. Is my work accurate, or do I need to do something else?
– Austin Brown
Nov 22 at 18:41
Yes, your formula is correct, R/L is the bandwidth in radians per second and that is 2pi more in value than in Hz so the error you have made is assuming that R/L equated to Hz.
– Andy aka
Nov 22 at 19:45
add a comment |
up vote
1
down vote
Where is the load, and what is the impedance of the load? Your filter will shift significantly based on what the output load actually is. Thus, you may want to buffer the output with a unity-gain opamp. Also, if your input isn't ideal (and it isn't,) it, too, may interact with your passive circuit, so you may need a buffer on that end, too.
Or you can design the entire circuit in context of its input/output stages, to figure out what the values should be as-used, if the input and output designs are fixed.
answered Nov 22 at 17:20
Jon Watte
4,7241534
1
I think this is homework and an "idealized" design situation. I believe the OP just needs to be able to demonstrate the ability to use Q and the value of some one part, perhaps, to calculate theoretical values.
– jonk
Nov 22 at 18:02
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
return StackExchange.using("schematics", function () {
StackExchange.schematics.init();
});
}, "cicuitlab");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "135"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f408273%2fdoes-this-look-like-a-correct-bandpass-filter%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The ratio of centre frequency to bandwidth is called Q and Q, for a series resonant circuit, is also dictated by R, L and C using this formula: -
$$Q = dfrac{1}{R}sqrt{dfrac{L}{C}}$$
Plugging in your circuit values I get a Q of 413. If I considered the centre-frequency to bandwidth ratio, I get 65.75 so, it looks like you will be "sharper" in resonance than your specification demands and this can cause problems.
I need to know if the method (using the equations I listed) is the correct way to determine my RLC values
The ratio of R to L does give you the bandwidth but that bandwidth is in radians per second; you have assumed it is in hertz hence, the value of Q of 413 (gleaned from my formula and your values) is precisely $2pi$ times higher than 65.7 ($f_0$ to bandwidth ratio) and that is your only mistake. The formula for resonant frequency is correct.
answered Nov 22 at 17:35
Andy aka
238k10174406
1
That's about what I get. I think this is totally a theoretical exercise for the OP and their Q seems excessive, just as you point out. I think this is the right kind of response, though.
– jonk
Nov 22 at 18:07
Yes this exercise is 100% theoretical.
– Austin Brown
Nov 22 at 18:37
I need to know if the method (using the equations I listed) is the correct way to determine my RLC values, and also that my graph is okay. Is my work accurate, or do I need to do something else?
– Austin Brown
Nov 22 at 18:41
Yes, your formula is correct, R/L is the bandwidth in radians per second and that is 2pi more in value than in Hz so the error you have made is assuming that R/L equated to Hz.
– Andy aka
Nov 22 at 19:45
add a comment |
up vote
2
down vote
The ratio of centre frequency to bandwidth is called Q and Q, for a series resonant circuit, is also dictated by R, L and C using this formula: -
$$Q = dfrac{1}{R}sqrt{dfrac{L}{C}}$$
Plugging in your circuit values I get a Q of 413. If I considered the centre-frequency to bandwidth ratio, I get 65.75 so, it looks like you will be "sharper" in resonance than your specification demands and this can cause problems.
I need to know if the method (using the equations I listed) is the correct way to determine my RLC values
The ratio of R to L does give you the bandwidth but that bandwidth is in radians per second; you have assumed it is in hertz hence, the value of Q of 413 (gleaned from my formula and your values) is precisely $2pi$ times higher than 65.7 ($f_0$ to bandwidth ratio) and that is your only mistake. The formula for resonant frequency is correct.
answered Nov 22 at 17:35
Andy aka
238k10174406
1
That's about what I get. I think this is totally a theoretical exercise for the OP and their Q seems excessive, just as you point out. I think this is the right kind of response, though.
– jonk
Nov 22 at 18:07
Yes this exercise is 100% theoretical.
– Austin Brown
Nov 22 at 18:37
I need to know if the method (using the equations I listed) is the correct way to determine my RLC values, and also that my graph is okay. Is my work accurate, or do I need to do something else?
– Austin Brown
Nov 22 at 18:41
Yes, your formula is correct, R/L is the bandwidth in radians per second and that is 2pi more in value than in Hz so the error you have made is assuming that R/L equated to Hz.
– Andy aka
Nov 22 at 19:45
add a comment |
up vote
2
down vote
up vote
2
down vote
The ratio of centre frequency to bandwidth is called Q and Q, for a series resonant circuit, is also dictated by R, L and C using this formula: -
$$Q = dfrac{1}{R}sqrt{dfrac{L}{C}}$$
Plugging in your circuit values I get a Q of 413. If I considered the centre-frequency to bandwidth ratio, I get 65.75 so, it looks like you will be "sharper" in resonance than your specification demands and this can cause problems.
I need to know if the method (using the equations I listed) is the correct way to determine my RLC values
The ratio of R to L does give you the bandwidth but that bandwidth is in radians per second; you have assumed it is in hertz hence, the value of Q of 413 (gleaned from my formula and your values) is precisely $2pi$ times higher than 65.7 ($f_0$ to bandwidth ratio) and that is your only mistake. The formula for resonant frequency is correct.
answered Nov 22 at 17:35
Andy aka
238k10174406
The ratio of centre frequency to bandwidth is called Q and Q, for a series resonant circuit, is also dictated by R, L and C using this formula: -
$$Q = dfrac{1}{R}sqrt{dfrac{L}{C}}$$
Plugging in your circuit values I get a Q of 413. If I considered the centre-frequency to bandwidth ratio, I get 65.75 so, it looks like you will be "sharper" in resonance than your specification demands and this can cause problems.
I need to know if the method (using the equations I listed) is the correct way to determine my RLC values
The ratio of R to L does give you the bandwidth but that bandwidth is in radians per second; you have assumed it is in hertz hence, the value of Q of 413 (gleaned from my formula and your values) is precisely $2pi$ times higher than 65.7 ($f_0$ to bandwidth ratio) and that is your only mistake. The formula for resonant frequency is correct.
answered Nov 22 at 17:35
Andy aka
238k10174406
edited Nov 22 at 19:53
answered Nov 22 at 17:35
Andy aka
238k10174406
answered Nov 22 at 17:35
Andy aka
238k10174406
answered Nov 22 at 17:35
Andy aka
238k10174406
238k10174406
1
That's about what I get. I think this is totally a theoretical exercise for the OP and their Q seems excessive, just as you point out. I think this is the right kind of response, though.
– jonk
Nov 22 at 18:07
Yes this exercise is 100% theoretical.
– Austin Brown
Nov 22 at 18:37
I need to know if the method (using the equations I listed) is the correct way to determine my RLC values, and also that my graph is okay. Is my work accurate, or do I need to do something else?
– Austin Brown
Nov 22 at 18:41
Yes, your formula is correct, R/L is the bandwidth in radians per second and that is 2pi more in value than in Hz so the error you have made is assuming that R/L equated to Hz.
– Andy aka
Nov 22 at 19:45
add a comment |
1
That's about what I get. I think this is totally a theoretical exercise for the OP and their Q seems excessive, just as you point out. I think this is the right kind of response, though.
– jonk
Nov 22 at 18:07
Yes this exercise is 100% theoretical.
– Austin Brown
Nov 22 at 18:37
I need to know if the method (using the equations I listed) is the correct way to determine my RLC values, and also that my graph is okay. Is my work accurate, or do I need to do something else?
– Austin Brown
Nov 22 at 18:41
Yes, your formula is correct, R/L is the bandwidth in radians per second and that is 2pi more in value than in Hz so the error you have made is assuming that R/L equated to Hz.
– Andy aka
Nov 22 at 19:45
1
1
That's about what I get. I think this is totally a theoretical exercise for the OP and their Q seems excessive, just as you point out. I think this is the right kind of response, though.
– jonk
Nov 22 at 18:07
That's about what I get. I think this is totally a theoretical exercise for the OP and their Q seems excessive, just as you point out. I think this is the right kind of response, though.
– jonk
Nov 22 at 18:07
Yes this exercise is 100% theoretical.
– Austin Brown
Nov 22 at 18:37
Yes this exercise is 100% theoretical.
– Austin Brown
Nov 22 at 18:37
I need to know if the method (using the equations I listed) is the correct way to determine my RLC values, and also that my graph is okay. Is my work accurate, or do I need to do something else?
– Austin Brown
Nov 22 at 18:41
I need to know if the method (using the equations I listed) is the correct way to determine my RLC values, and also that my graph is okay. Is my work accurate, or do I need to do something else?
– Austin Brown
Nov 22 at 18:41
Yes, your formula is correct, R/L is the bandwidth in radians per second and that is 2pi more in value than in Hz so the error you have made is assuming that R/L equated to Hz.
– Andy aka
Nov 22 at 19:45
Yes, your formula is correct, R/L is the bandwidth in radians per second and that is 2pi more in value than in Hz so the error you have made is assuming that R/L equated to Hz.
– Andy aka
Nov 22 at 19:45
add a comment |
up vote
1
down vote
Where is the load, and what is the impedance of the load? Your filter will shift significantly based on what the output load actually is. Thus, you may want to buffer the output with a unity-gain opamp. Also, if your input isn't ideal (and it isn't,) it, too, may interact with your passive circuit, so you may need a buffer on that end, too.
Or you can design the entire circuit in context of its input/output stages, to figure out what the values should be as-used, if the input and output designs are fixed.
answered Nov 22 at 17:20
Jon Watte
4,7241534
1
I think this is homework and an "idealized" design situation. I believe the OP just needs to be able to demonstrate the ability to use Q and the value of some one part, perhaps, to calculate theoretical values.
– jonk
Nov 22 at 18:02
add a comment |
up vote
1
down vote
Where is the load, and what is the impedance of the load? Your filter will shift significantly based on what the output load actually is. Thus, you may want to buffer the output with a unity-gain opamp. Also, if your input isn't ideal (and it isn't,) it, too, may interact with your passive circuit, so you may need a buffer on that end, too.
Or you can design the entire circuit in context of its input/output stages, to figure out what the values should be as-used, if the input and output designs are fixed.
answered Nov 22 at 17:20
Jon Watte
4,7241534
1
I think this is homework and an "idealized" design situation. I believe the OP just needs to be able to demonstrate the ability to use Q and the value of some one part, perhaps, to calculate theoretical values.
– jonk
Nov 22 at 18:02
add a comment |
up vote
1
down vote
up vote
1
down vote
Where is the load, and what is the impedance of the load? Your filter will shift significantly based on what the output load actually is. Thus, you may want to buffer the output with a unity-gain opamp. Also, if your input isn't ideal (and it isn't,) it, too, may interact with your passive circuit, so you may need a buffer on that end, too.
Or you can design the entire circuit in context of its input/output stages, to figure out what the values should be as-used, if the input and output designs are fixed.
answered Nov 22 at 17:20
Jon Watte
4,7241534
Where is the load, and what is the impedance of the load? Your filter will shift significantly based on what the output load actually is. Thus, you may want to buffer the output with a unity-gain opamp. Also, if your input isn't ideal (and it isn't,) it, too, may interact with your passive circuit, so you may need a buffer on that end, too.
Or you can design the entire circuit in context of its input/output stages, to figure out what the values should be as-used, if the input and output designs are fixed.
answered Nov 22 at 17:20
Jon Watte
4,7241534
answered Nov 22 at 17:20
Jon Watte
4,7241534
answered Nov 22 at 17:20
Jon Watte
4,7241534
answered Nov 22 at 17:20
Jon Watte
4,7241534
4,7241534
1
I think this is homework and an "idealized" design situation. I believe the OP just needs to be able to demonstrate the ability to use Q and the value of some one part, perhaps, to calculate theoretical values.
– jonk
Nov 22 at 18:02
add a comment |
1
I think this is homework and an "idealized" design situation. I believe the OP just needs to be able to demonstrate the ability to use Q and the value of some one part, perhaps, to calculate theoretical values.
– jonk
Nov 22 at 18:02
1
1
I think this is homework and an "idealized" design situation. I believe the OP just needs to be able to demonstrate the ability to use Q and the value of some one part, perhaps, to calculate theoretical values.
– jonk
Nov 22 at 18:02
I think this is homework and an "idealized" design situation. I believe the OP just needs to be able to demonstrate the ability to use Q and the value of some one part, perhaps, to calculate theoretical values.
– jonk
Nov 22 at 18:02
add a comment |
Thanks for contributing an answer to Electrical Engineering Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f408273%2fdoes-this-look-like-a-correct-bandpass-filter%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Set your x-axis to a log scale and you'll see something more like what you would see in a textbook.
– John D
Nov 22 at 17:20
I would have gone with $C_1=100:text{pF}$ as a starting value and used Q and the center frequency's relationship to LC to work out the values. However, the values you have "feel" they are in the right ballpark. Your Q seems a bit high. But that's easily fixed.
– jonk
Nov 22 at 18:06
To verify if your bandwidth is exactly 73kHz, expand your x scale (don't make it a log scale) and look for the points where the response is 3dB down from the peak. They should be 73kHz apart.
– TimWescott
Nov 22 at 18:28
@TimWescott I tested this and this gives me 12kHz apart only. Why is it that you choose 3dB? :)
– Austin Brown
Nov 22 at 18:36
I agree with @TimWescott about not making it a log scale when making BW measurements, I was just suggesting the log scale because many textbook bandpass curves are plotted on a log scale and the shape looks more familiar that way. (In case that was one of the concerns with the question "Is the graph what I should expect?"
– John D
Nov 22 at 19:00