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Pole-zeros of a real-valued causal FIR system
The Next CEO of Stack OverflowPoles 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
asked 2 days ago
NioushaNiousha
1596
$endgroup$
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
asked 2 days ago
NioushaNiousha
1596
$endgroup$
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
asked 2 days ago
NioushaNiousha
1596
$endgroup$
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 2 days ago
NioushaNiousha
1596
asked 2 days ago
NioushaNiousha
1596
asked 2 days ago
NioushaNiousha
1596
asked 2 days ago
NioushaNiousha
1596
asked 2 days ago
NioushaNiousha
1596
1596
add a comment |
add a comment |
2 Answers
2
active
oldest
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 $z = e^pm j 0.8 pi$
answered 2 days ago
JuanchoJuancho
3,8801315
$endgroup$
add a comment |
$begingroup$
Causality places the transfer-function poles at $z=0$, for a FIR filter.
A FIR filter need not necessarily be causal, in which case some or all of its poles reside at $z=infty$ (if not at $z=0$). In any of these cases, the poles play no role in shaping frequency response, since they remain equidistant from the unit circle.
(A value of $k$ in the range of $[-1, 1]$ can place conjugate pole pairs anywhere on the unit circle, where they are most effectual in shaping frequency response.)
$$beginaligned
fracz^2 +2kz + 1z^2 &mboximplies y_n=x_n + 2kx_n-1 + x_n-2 \
z^2 +2kz + 1 &mboximplies y_n=x_n+2 + 2kx_n+1 + x_n
endaligned$$
answered yesterday
KevinKevin
111
New contributor
Kevin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
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 $z = e^pm j 0.8 pi$
answered 2 days ago
JuanchoJuancho
3,8801315
$endgroup$
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 $z = e^pm j 0.8 pi$
answered 2 days ago
JuanchoJuancho
3,8801315
$endgroup$
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 $z = e^pm j 0.8 pi$
answered 2 days ago
JuanchoJuancho
3,8801315
$endgroup$
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 2 days ago
JuanchoJuancho
3,8801315
answered 2 days ago
JuanchoJuancho
3,8801315
answered 2 days ago
JuanchoJuancho
3,8801315
answered 2 days ago
JuanchoJuancho
3,8801315
3,8801315
add a comment |
add a comment |
$begingroup$
Causality places the transfer-function poles at $z=0$, for a FIR filter.
A FIR filter need not necessarily be causal, in which case some or all of its poles reside at $z=infty$ (if not at $z=0$). In any of these cases, the poles play no role in shaping frequency response, since they remain equidistant from the unit circle.
(A value of $k$ in the range of $[-1, 1]$ can place conjugate pole pairs anywhere on the unit circle, where they are most effectual in shaping frequency response.)
$$beginaligned
fracz^2 +2kz + 1z^2 &mboximplies y_n=x_n + 2kx_n-1 + x_n-2 \
z^2 +2kz + 1 &mboximplies y_n=x_n+2 + 2kx_n+1 + x_n
endaligned$$
answered yesterday
KevinKevin
111
New contributor
Kevin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Causality places the transfer-function poles at $z=0$, for a FIR filter.
A FIR filter need not necessarily be causal, in which case some or all of its poles reside at $z=infty$ (if not at $z=0$). In any of these cases, the poles play no role in shaping frequency response, since they remain equidistant from the unit circle.
(A value of $k$ in the range of $[-1, 1]$ can place conjugate pole pairs anywhere on the unit circle, where they are most effectual in shaping frequency response.)
$$beginaligned
fracz^2 +2kz + 1z^2 &mboximplies y_n=x_n + 2kx_n-1 + x_n-2 \
z^2 +2kz + 1 &mboximplies y_n=x_n+2 + 2kx_n+1 + x_n
endaligned$$
answered yesterday
KevinKevin
111
New contributor
Kevin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Causality places the transfer-function poles at $z=0$, for a FIR filter.
A FIR filter need not necessarily be causal, in which case some or all of its poles reside at $z=infty$ (if not at $z=0$). In any of these cases, the poles play no role in shaping frequency response, since they remain equidistant from the unit circle.
(A value of $k$ in the range of $[-1, 1]$ can place conjugate pole pairs anywhere on the unit circle, where they are most effectual in shaping frequency response.)
$$beginaligned
fracz^2 +2kz + 1z^2 &mboximplies y_n=x_n + 2kx_n-1 + x_n-2 \
z^2 +2kz + 1 &mboximplies y_n=x_n+2 + 2kx_n+1 + x_n
endaligned$$
answered yesterday
KevinKevin
111
New contributor
Kevin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Causality places the transfer-function poles at $z=0$, for a FIR filter.
A FIR filter need not necessarily be causal, in which case some or all of its poles reside at $z=infty$ (if not at $z=0$). In any of these cases, the poles play no role in shaping frequency response, since they remain equidistant from the unit circle.
(A value of $k$ in the range of $[-1, 1]$ can place conjugate pole pairs anywhere on the unit circle, where they are most effectual in shaping frequency response.)
$$beginaligned
fracz^2 +2kz + 1z^2 &mboximplies y_n=x_n + 2kx_n-1 + x_n-2 \
z^2 +2kz + 1 &mboximplies y_n=x_n+2 + 2kx_n+1 + x_n
endaligned$$
answered yesterday
KevinKevin
111
New contributor
Kevin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
KevinKevin
111
New contributor
Kevin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
KevinKevin
111
answered yesterday
KevinKevin
111
111
New contributor
Kevin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kevin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kevin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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