Pole-zeros of a real-valued causal FIR system
$begingroup$
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 Mar 27 at 19:40
NioushaNiousha
1646
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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 Mar 27 at 19:40
NioushaNiousha
1646
$endgroup$
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 Mar 27 at 19:40
NioushaNiousha
1646
$endgroup$
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 Mar 27 at 19:40
NioushaNiousha
1646
asked Mar 27 at 19:40
NioushaNiousha
1646
asked Mar 27 at 19:40
NioushaNiousha
1646
asked Mar 27 at 19:40
NioushaNiousha
1646
asked Mar 27 at 19:40
NioushaNiousha
1646
1646
add a comment |
add a comment |
2 Answers
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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 Mar 27 at 20:10
JuanchoJuancho
3,8901416
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$begingroup$
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.)
$$begin{aligned}
frac{z^2 +2kz + 1}{z^2} &mbox{implies} y_n=x_n + 2kx_{n-1} + x_{n-2} \
z^2 +2kz + 1 &mbox{implies} y_n=x_{n+2} + 2kx_{n+1} + x_{n}
end{aligned}$$
answered Mar 28 at 7:32
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.
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2 Answers
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2 Answers
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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 $z = e^{pm j 0.8 pi}$
answered Mar 27 at 20:10
JuanchoJuancho
3,8901416
$endgroup$
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 Mar 27 at 20:10
JuanchoJuancho
3,8901416
$endgroup$
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 Mar 27 at 20:10
JuanchoJuancho
3,8901416
$endgroup$
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 Mar 27 at 20:10
JuanchoJuancho
3,8901416
answered Mar 27 at 20:10
JuanchoJuancho
3,8901416
answered Mar 27 at 20:10
JuanchoJuancho
3,8901416
answered Mar 27 at 20:10
JuanchoJuancho
3,8901416
3,8901416
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.)
$$begin{aligned}
frac{z^2 +2kz + 1}{z^2} &mbox{implies} y_n=x_n + 2kx_{n-1} + x_{n-2} \
z^2 +2kz + 1 &mbox{implies} y_n=x_{n+2} + 2kx_{n+1} + x_{n}
end{aligned}$$
answered Mar 28 at 7:32
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.)
$$begin{aligned}
frac{z^2 +2kz + 1}{z^2} &mbox{implies} y_n=x_n + 2kx_{n-1} + x_{n-2} \
z^2 +2kz + 1 &mbox{implies} y_n=x_{n+2} + 2kx_{n+1} + x_{n}
end{aligned}$$
answered Mar 28 at 7:32
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.)
$$begin{aligned}
frac{z^2 +2kz + 1}{z^2} &mbox{implies} y_n=x_n + 2kx_{n-1} + x_{n-2} \
z^2 +2kz + 1 &mbox{implies} y_n=x_{n+2} + 2kx_{n+1} + x_{n}
end{aligned}$$
answered Mar 28 at 7:32
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.)
$$begin{aligned}
frac{z^2 +2kz + 1}{z^2} &mbox{implies} y_n=x_n + 2kx_{n-1} + x_{n-2} \
z^2 +2kz + 1 &mbox{implies} y_n=x_{n+2} + 2kx_{n+1} + x_{n}
end{aligned}$$
answered Mar 28 at 7:32
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 Mar 28 at 7:32
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 Mar 28 at 7:32
KevinKevin
111
answered Mar 28 at 7:32
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.
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