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Pole-zeros of a real-valued causal FIR system
Poles and Zerospole/zero locations for real and imaginary signalIdentifying the magnitude and impulse response from pole zero plot quicklySystem characterization given pole-zero mappingWhat's the Q of a pole at the origin of the s-plane?How to find system function, H(z) in the z-domain, given zero-pole plot of the system?Conjugate Pole PairsQuestion about poles and zeros in AR filterDetermine poles and zeros of a specific filter designHow to determine if a filter is bandpass/stopband from its pole-zero diagram in z-domain
$begingroup$
Could someone please help me with the following question?
Below is the magnitude response of a real-valued causal linear phase FIR system of order N = 6. Determine the location of poles and zeros.
I know that for FIR systems all the poles are located at the origin, so we have a pole of order six at the origin. Also from the given diagram, I can say that we have a zero at 0.3pi and one at 0.8pi (both on the unit circle). Now since the system is real-valued, location of poles and zeros should be symmetric w.r.t. the real axis. But I don't know about the two other zeros?
Also, what about the pick in the diagram? Does it mean we have another pole?
fir poles-zeros
$endgroup$
add a comment |
$begingroup$
Could someone please help me with the following question?
Below is the magnitude response of a real-valued causal linear phase FIR system of order N = 6. Determine the location of poles and zeros.
I know that for FIR systems all the poles are located at the origin, so we have a pole of order six at the origin. Also from the given diagram, I can say that we have a zero at 0.3pi and one at 0.8pi (both on the unit circle). Now since the system is real-valued, location of poles and zeros should be symmetric w.r.t. the real axis. But I don't know about the two other zeros?
Also, what about the pick in the diagram? Does it mean we have another pole?
fir poles-zeros
$endgroup$
add a comment |
$begingroup$
Could someone please help me with the following question?
Below is the magnitude response of a real-valued causal linear phase FIR system of order N = 6. Determine the location of poles and zeros.
I know that for FIR systems all the poles are located at the origin, so we have a pole of order six at the origin. Also from the given diagram, I can say that we have a zero at 0.3pi and one at 0.8pi (both on the unit circle). Now since the system is real-valued, location of poles and zeros should be symmetric w.r.t. the real axis. But I don't know about the two other zeros?
Also, what about the pick in the diagram? Does it mean we have another pole?
fir poles-zeros
$endgroup$
Could someone please help me with the following question?
Below is the magnitude response of a real-valued causal linear phase FIR system of order N = 6. Determine the location of poles and zeros.
I know that for FIR systems all the poles are located at the origin, so we have a pole of order six at the origin. Also from the given diagram, I can say that we have a zero at 0.3pi and one at 0.8pi (both on the unit circle). Now since the system is real-valued, location of poles and zeros should be symmetric w.r.t. the real axis. But I don't know about the two other zeros?
Also, what about the pick in the diagram? Does it mean we have another pole?
fir poles-zeros
fir poles-zeros
asked 9 hours ago
NioushaNiousha
1346
1346
add a comment |
add a comment |
1 Answer
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$begingroup$
Note the difference between the zeros at $0.3 pi$ and at $0.8 pi$.
The first one is clearly a zero crossing, much like $abs(x)$ at $x=0$.
At $theta = 0.8 pi$, however, the curve is tangent to the horizontal axis, much like $x^2$ at $x=0$. So you have a doulbe zero here.
So your zeros are:
- 2 zeros at $z = e^pm j 0.3 pi$
- 2 double zeros at $z = e^pm j 0.8 pi$
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
Note the difference between the zeros at $0.3 pi$ and at $0.8 pi$.
The first one is clearly a zero crossing, much like $abs(x)$ at $x=0$.
At $theta = 0.8 pi$, however, the curve is tangent to the horizontal axis, much like $x^2$ at $x=0$. So you have a doulbe zero here.
So your zeros are:
- 2 zeros at $z = e^pm j 0.3 pi$
- 2 double zeros at $z = e^pm j 0.8 pi$
$endgroup$
add a comment |
$begingroup$
Note the difference between the zeros at $0.3 pi$ and at $0.8 pi$.
The first one is clearly a zero crossing, much like $abs(x)$ at $x=0$.
At $theta = 0.8 pi$, however, the curve is tangent to the horizontal axis, much like $x^2$ at $x=0$. So you have a doulbe zero here.
So your zeros are:
- 2 zeros at $z = e^pm j 0.3 pi$
- 2 double zeros at $z = e^pm j 0.8 pi$
$endgroup$
add a comment |
$begingroup$
Note the difference between the zeros at $0.3 pi$ and at $0.8 pi$.
The first one is clearly a zero crossing, much like $abs(x)$ at $x=0$.
At $theta = 0.8 pi$, however, the curve is tangent to the horizontal axis, much like $x^2$ at $x=0$. So you have a doulbe zero here.
So your zeros are:
- 2 zeros at $z = e^pm j 0.3 pi$
- 2 double zeros at $z = e^pm j 0.8 pi$
$endgroup$
Note the difference between the zeros at $0.3 pi$ and at $0.8 pi$.
The first one is clearly a zero crossing, much like $abs(x)$ at $x=0$.
At $theta = 0.8 pi$, however, the curve is tangent to the horizontal axis, much like $x^2$ at $x=0$. So you have a doulbe zero here.
So your zeros are:
- 2 zeros at $z = e^pm j 0.3 pi$
- 2 double zeros at $z = e^pm j 0.8 pi$
answered 9 hours ago
JuanchoJuancho
3,8501214
3,8501214
add a comment |
add a comment |
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