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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

5

1

$begingroup$

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

share|improve this question

asked Mar 27 at 19:40

NioushaNiousha

1546

$endgroup$

add a comment | 

5

1

$begingroup$

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

share|improve this question

asked Mar 27 at 19:40

NioushaNiousha

1546

$endgroup$

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

share|improve this question

asked Mar 27 at 19:40

NioushaNiousha

1546

$endgroup$

share|improve this question

asked Mar 27 at 19:40

NioushaNiousha

1546

share|improve this question

asked Mar 27 at 19:40

NioushaNiousha

1546

asked Mar 27 at 19:40

NioushaNiousha

1546

asked Mar 27 at 19:40

NioushaNiousha

1546

2 Answers
2

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10

$begingroup$

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 $z = e^pm j 0.8 pi$
  

share|improve this answer

answered Mar 27 at 20:10

JuanchoJuancho

3,8901416

$endgroup$

add a comment | 

1

$begingroup$

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.)
$$beginaligned
fracz^2 +2kz + 1z^2 &mboximplies y_n=x_n + 2kx_n-1 + x_n-2 \
z^2 +2kz + 1 &mboximplies y_n=x_n+2 + 2kx_n+1 + x_n
endaligned$$

share|improve this answer

answered Mar 28 at 7:32

KevinKevin

111

$endgroup$

add a comment | 

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10

$begingroup$

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 $z = e^pm j 0.8 pi$
  

share|improve this answer

answered Mar 27 at 20:10

JuanchoJuancho

3,8901416

$endgroup$

add a comment | 

10

$begingroup$

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 $z = e^pm j 0.8 pi$
  

share|improve this answer

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 $z = e^pm j 0.8 pi$
  

share|improve this answer

answered Mar 27 at 20:10

JuanchoJuancho

3,8901416

$endgroup$

share|improve this answer

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

1

$begingroup$

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.)
$$beginaligned
fracz^2 +2kz + 1z^2 &mboximplies y_n=x_n + 2kx_n-1 + x_n-2 \
z^2 +2kz + 1 &mboximplies y_n=x_n+2 + 2kx_n+1 + x_n
endaligned$$

share|improve this answer

answered Mar 28 at 7:32

KevinKevin

111

$endgroup$

add a comment | 

1

$begingroup$

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.)
$$beginaligned
fracz^2 +2kz + 1z^2 &mboximplies y_n=x_n + 2kx_n-1 + x_n-2 \
z^2 +2kz + 1 &mboximplies y_n=x_n+2 + 2kx_n+1 + x_n
endaligned$$

share|improve this answer

answered Mar 28 at 7:32

KevinKevin

111

$endgroup$

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.)
$$beginaligned
fracz^2 +2kz + 1z^2 &mboximplies y_n=x_n + 2kx_n-1 + x_n-2 \
z^2 +2kz + 1 &mboximplies y_n=x_n+2 + 2kx_n+1 + x_n
endaligned$$

share|improve this answer

answered Mar 28 at 7:32

KevinKevin

111

$endgroup$

share|improve this answer

answered Mar 28 at 7:32

KevinKevin

111

answered Mar 28 at 7:32

KevinKevin

111

answered Mar 28 at 7:32

KevinKevin

111