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Z Transform - Do i always need 2 poles for every “peak�
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I am quite new to digital signal processing (and also Z transforms). I am reading about frequency response modelling and have some questions
- do we always need a pair of poles for each "peak" in the frequency response ?
- why is this so ?
- what happens if i dont have a pair of pole for each peak ?
frequency-spectrum z-transform frequency-response
asked 4 hours ago
kong
1253
add a comment |Â
up vote
1
down vote
favorite
I am quite new to digital signal processing (and also Z transforms). I am reading about frequency response modelling and have some questions
- do we always need a pair of poles for each "peak" in the frequency response ?
- why is this so ?
- what happens if i dont have a pair of pole for each peak ?
frequency-spectrum z-transform frequency-response
asked 4 hours ago
kong
1253
Please clarify what you mean by “peakâ€Â; do you mean a frequency when the frequency response goes to infinity or is just at a maximum? What about having a peak at DC?
– Dan Boschen
4 hours ago
Also is your thinking restricted to only real inputs (sinusoids) or are you able to understand yet what a complex exponential frequency is? (Since you said this was all new to you)
– Dan Boschen
4 hours ago
@DanBoschen I meant when the frequency goes to infinity. Does having a peak at DC or not change anything ? I know Euler's formula.. is that called a complex exponential frequency ?
– kong
4 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am quite new to digital signal processing (and also Z transforms). I am reading about frequency response modelling and have some questions
- do we always need a pair of poles for each "peak" in the frequency response ?
- why is this so ?
- what happens if i dont have a pair of pole for each peak ?
frequency-spectrum z-transform frequency-response
asked 4 hours ago
kong
1253
I am quite new to digital signal processing (and also Z transforms). I am reading about frequency response modelling and have some questions
- do we always need a pair of poles for each "peak" in the frequency response ?
- why is this so ?
- what happens if i dont have a pair of pole for each peak ?
frequency-spectrum z-transform frequency-response
frequency-spectrum z-transform frequency-response
asked 4 hours ago
kong
1253
asked 4 hours ago
kong
1253
asked 4 hours ago
kong
1253
asked 4 hours ago
kong
1253
asked 4 hours ago
kong
1253
1253
Please clarify what you mean by “peakâ€Â; do you mean a frequency when the frequency response goes to infinity or is just at a maximum? What about having a peak at DC?
– Dan Boschen
4 hours ago
Also is your thinking restricted to only real inputs (sinusoids) or are you able to understand yet what a complex exponential frequency is? (Since you said this was all new to you)
– Dan Boschen
4 hours ago
@DanBoschen I meant when the frequency goes to infinity. Does having a peak at DC or not change anything ? I know Euler's formula.. is that called a complex exponential frequency ?
– kong
4 hours ago
add a comment |Â
Please clarify what you mean by “peakâ€Â; do you mean a frequency when the frequency response goes to infinity or is just at a maximum? What about having a peak at DC?
– Dan Boschen
4 hours ago
Also is your thinking restricted to only real inputs (sinusoids) or are you able to understand yet what a complex exponential frequency is? (Since you said this was all new to you)
– Dan Boschen
4 hours ago
@DanBoschen I meant when the frequency goes to infinity. Does having a peak at DC or not change anything ? I know Euler's formula.. is that called a complex exponential frequency ?
– kong
4 hours ago
Please clarify what you mean by “peakâ€Â; do you mean a frequency when the frequency response goes to infinity or is just at a maximum? What about having a peak at DC?
– Dan Boschen
4 hours ago
Please clarify what you mean by “peakâ€Â; do you mean a frequency when the frequency response goes to infinity or is just at a maximum? What about having a peak at DC?
– Dan Boschen
4 hours ago
Also is your thinking restricted to only real inputs (sinusoids) or are you able to understand yet what a complex exponential frequency is? (Since you said this was all new to you)
– Dan Boschen
4 hours ago
Also is your thinking restricted to only real inputs (sinusoids) or are you able to understand yet what a complex exponential frequency is? (Since you said this was all new to you)
– Dan Boschen
4 hours ago
@DanBoschen I meant when the frequency goes to infinity. Does having a peak at DC or not change anything ? I know Euler's formula.. is that called a complex exponential frequency ?
– kong
4 hours ago
@DanBoschen I meant when the frequency goes to infinity. Does having a peak at DC or not change anything ? I know Euler's formula.. is that called a complex exponential frequency ?
– kong
4 hours ago
add a comment |Â
2 Answers
2
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oldest
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up vote
1
down vote
accepted
If your system has a real impulse response, then pole & zeros will be in complex-conjugate pairs; so a pole at $omega = pi/3$ implies another pole at $omega = -pi/3$ .
Further, by the definition of poles, as the frequency $omega$ get closer to the pole frequency, the frequency response magnitude increases (so as to say peaks!). So for every peak in the (positive frequency axis) then you would have at least two poles at the same frequency (one for positive frequency and one for the negative frequency). Note that there could be more, but at least two is required for a real system.
However note that, having $2N$ poles does not necessarily mean that the frequency response curve has $N$ peaks (sharp rising transitions). Indeed counter-examples are easiy to find, such as an Butterworth or elliptic IIR lowpass filters having no peaks, compared to a sharp isolated rising transition.
Note that in those filters the poles are bundled together such that their indivual peaks have overlapping effects on the total frequency response (yielding a flat passband rather than isolated peaks).
answered 3 hours ago
Fat32
12.2k31126
add a comment |Â
up vote
2
down vote
To answer your question specifically, no you do not need two poles. A single pole anywhere on the frequency axis will cause a real or complex frequency input to go to infinity (peak). For a digital system, the frequency axis is the unit circle. As Fat32 very well summarized, if your impulse response (which the poles and zeros are the z-transform or Laplace transform of for digital or an analog system specifically) is real, then complex conjugate poles are required. So in the case of DC (or fs/2 for a digital system where fs is the sampling rate) only one pole can exist in these cases even with a real system, otherwise for any other real system with a single frequency peak, 2 poles must exist.
answered 3 hours ago
Dan Boschen
8,7582933
Thank you. But it looks like I lack the understanding of something very simple: real vs imaginary systems/transfer functions and what they mean... because now I don't understand why complex conjugate poles are required for an impulse response that is real. What happens if I don't have 2 poles for an impulse response that is real ?
– kong
2 hours ago
1
@kong that comes from the fact that a polynomial of real coefficients can only have real roots or complex roots in conjugate pairs. That polynomial is the characteristic equation polynomial of the LCCDE equation associated with the real system, which has real coefficients by definition.
– Fat32
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If your system has a real impulse response, then pole & zeros will be in complex-conjugate pairs; so a pole at $omega = pi/3$ implies another pole at $omega = -pi/3$ .
Further, by the definition of poles, as the frequency $omega$ get closer to the pole frequency, the frequency response magnitude increases (so as to say peaks!). So for every peak in the (positive frequency axis) then you would have at least two poles at the same frequency (one for positive frequency and one for the negative frequency). Note that there could be more, but at least two is required for a real system.
However note that, having $2N$ poles does not necessarily mean that the frequency response curve has $N$ peaks (sharp rising transitions). Indeed counter-examples are easiy to find, such as an Butterworth or elliptic IIR lowpass filters having no peaks, compared to a sharp isolated rising transition.
Note that in those filters the poles are bundled together such that their indivual peaks have overlapping effects on the total frequency response (yielding a flat passband rather than isolated peaks).
answered 3 hours ago
Fat32
12.2k31126
add a comment |Â
up vote
1
down vote
accepted
If your system has a real impulse response, then pole & zeros will be in complex-conjugate pairs; so a pole at $omega = pi/3$ implies another pole at $omega = -pi/3$ .
Further, by the definition of poles, as the frequency $omega$ get closer to the pole frequency, the frequency response magnitude increases (so as to say peaks!). So for every peak in the (positive frequency axis) then you would have at least two poles at the same frequency (one for positive frequency and one for the negative frequency). Note that there could be more, but at least two is required for a real system.
However note that, having $2N$ poles does not necessarily mean that the frequency response curve has $N$ peaks (sharp rising transitions). Indeed counter-examples are easiy to find, such as an Butterworth or elliptic IIR lowpass filters having no peaks, compared to a sharp isolated rising transition.
Note that in those filters the poles are bundled together such that their indivual peaks have overlapping effects on the total frequency response (yielding a flat passband rather than isolated peaks).
answered 3 hours ago
Fat32
12.2k31126
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If your system has a real impulse response, then pole & zeros will be in complex-conjugate pairs; so a pole at $omega = pi/3$ implies another pole at $omega = -pi/3$ .
Further, by the definition of poles, as the frequency $omega$ get closer to the pole frequency, the frequency response magnitude increases (so as to say peaks!). So for every peak in the (positive frequency axis) then you would have at least two poles at the same frequency (one for positive frequency and one for the negative frequency). Note that there could be more, but at least two is required for a real system.
However note that, having $2N$ poles does not necessarily mean that the frequency response curve has $N$ peaks (sharp rising transitions). Indeed counter-examples are easiy to find, such as an Butterworth or elliptic IIR lowpass filters having no peaks, compared to a sharp isolated rising transition.
Note that in those filters the poles are bundled together such that their indivual peaks have overlapping effects on the total frequency response (yielding a flat passband rather than isolated peaks).
answered 3 hours ago
Fat32
12.2k31126
If your system has a real impulse response, then pole & zeros will be in complex-conjugate pairs; so a pole at $omega = pi/3$ implies another pole at $omega = -pi/3$ .
Further, by the definition of poles, as the frequency $omega$ get closer to the pole frequency, the frequency response magnitude increases (so as to say peaks!). So for every peak in the (positive frequency axis) then you would have at least two poles at the same frequency (one for positive frequency and one for the negative frequency). Note that there could be more, but at least two is required for a real system.
However note that, having $2N$ poles does not necessarily mean that the frequency response curve has $N$ peaks (sharp rising transitions). Indeed counter-examples are easiy to find, such as an Butterworth or elliptic IIR lowpass filters having no peaks, compared to a sharp isolated rising transition.
Note that in those filters the poles are bundled together such that their indivual peaks have overlapping effects on the total frequency response (yielding a flat passband rather than isolated peaks).
answered 3 hours ago
Fat32
12.2k31126
answered 3 hours ago
Fat32
12.2k31126
answered 3 hours ago
Fat32
12.2k31126
answered 3 hours ago
Fat32
12.2k31126
12.2k31126
add a comment |Â
add a comment |Â
up vote
2
down vote
To answer your question specifically, no you do not need two poles. A single pole anywhere on the frequency axis will cause a real or complex frequency input to go to infinity (peak). For a digital system, the frequency axis is the unit circle. As Fat32 very well summarized, if your impulse response (which the poles and zeros are the z-transform or Laplace transform of for digital or an analog system specifically) is real, then complex conjugate poles are required. So in the case of DC (or fs/2 for a digital system where fs is the sampling rate) only one pole can exist in these cases even with a real system, otherwise for any other real system with a single frequency peak, 2 poles must exist.
answered 3 hours ago
Dan Boschen
8,7582933
Thank you. But it looks like I lack the understanding of something very simple: real vs imaginary systems/transfer functions and what they mean... because now I don't understand why complex conjugate poles are required for an impulse response that is real. What happens if I don't have 2 poles for an impulse response that is real ?
– kong
2 hours ago
1
@kong that comes from the fact that a polynomial of real coefficients can only have real roots or complex roots in conjugate pairs. That polynomial is the characteristic equation polynomial of the LCCDE equation associated with the real system, which has real coefficients by definition.
– Fat32
1 hour ago
add a comment |Â
up vote
2
down vote
To answer your question specifically, no you do not need two poles. A single pole anywhere on the frequency axis will cause a real or complex frequency input to go to infinity (peak). For a digital system, the frequency axis is the unit circle. As Fat32 very well summarized, if your impulse response (which the poles and zeros are the z-transform or Laplace transform of for digital or an analog system specifically) is real, then complex conjugate poles are required. So in the case of DC (or fs/2 for a digital system where fs is the sampling rate) only one pole can exist in these cases even with a real system, otherwise for any other real system with a single frequency peak, 2 poles must exist.
answered 3 hours ago
Dan Boschen
8,7582933
Thank you. But it looks like I lack the understanding of something very simple: real vs imaginary systems/transfer functions and what they mean... because now I don't understand why complex conjugate poles are required for an impulse response that is real. What happens if I don't have 2 poles for an impulse response that is real ?
– kong
2 hours ago
1
@kong that comes from the fact that a polynomial of real coefficients can only have real roots or complex roots in conjugate pairs. That polynomial is the characteristic equation polynomial of the LCCDE equation associated with the real system, which has real coefficients by definition.
– Fat32
1 hour ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
To answer your question specifically, no you do not need two poles. A single pole anywhere on the frequency axis will cause a real or complex frequency input to go to infinity (peak). For a digital system, the frequency axis is the unit circle. As Fat32 very well summarized, if your impulse response (which the poles and zeros are the z-transform or Laplace transform of for digital or an analog system specifically) is real, then complex conjugate poles are required. So in the case of DC (or fs/2 for a digital system where fs is the sampling rate) only one pole can exist in these cases even with a real system, otherwise for any other real system with a single frequency peak, 2 poles must exist.
answered 3 hours ago
Dan Boschen
8,7582933
To answer your question specifically, no you do not need two poles. A single pole anywhere on the frequency axis will cause a real or complex frequency input to go to infinity (peak). For a digital system, the frequency axis is the unit circle. As Fat32 very well summarized, if your impulse response (which the poles and zeros are the z-transform or Laplace transform of for digital or an analog system specifically) is real, then complex conjugate poles are required. So in the case of DC (or fs/2 for a digital system where fs is the sampling rate) only one pole can exist in these cases even with a real system, otherwise for any other real system with a single frequency peak, 2 poles must exist.
answered 3 hours ago
Dan Boschen
8,7582933
edited 2 hours ago
answered 3 hours ago
Dan Boschen
8,7582933
answered 3 hours ago
Dan Boschen
8,7582933
answered 3 hours ago
Dan Boschen
8,7582933
8,7582933
Thank you. But it looks like I lack the understanding of something very simple: real vs imaginary systems/transfer functions and what they mean... because now I don't understand why complex conjugate poles are required for an impulse response that is real. What happens if I don't have 2 poles for an impulse response that is real ?
– kong
2 hours ago
1
@kong that comes from the fact that a polynomial of real coefficients can only have real roots or complex roots in conjugate pairs. That polynomial is the characteristic equation polynomial of the LCCDE equation associated with the real system, which has real coefficients by definition.
– Fat32
1 hour ago
add a comment |Â
Thank you. But it looks like I lack the understanding of something very simple: real vs imaginary systems/transfer functions and what they mean... because now I don't understand why complex conjugate poles are required for an impulse response that is real. What happens if I don't have 2 poles for an impulse response that is real ?
– kong
2 hours ago
1
@kong that comes from the fact that a polynomial of real coefficients can only have real roots or complex roots in conjugate pairs. That polynomial is the characteristic equation polynomial of the LCCDE equation associated with the real system, which has real coefficients by definition.
– Fat32
1 hour ago
Thank you. But it looks like I lack the understanding of something very simple: real vs imaginary systems/transfer functions and what they mean... because now I don't understand why complex conjugate poles are required for an impulse response that is real. What happens if I don't have 2 poles for an impulse response that is real ?
– kong
2 hours ago
Thank you. But it looks like I lack the understanding of something very simple: real vs imaginary systems/transfer functions and what they mean... because now I don't understand why complex conjugate poles are required for an impulse response that is real. What happens if I don't have 2 poles for an impulse response that is real ?
– kong
2 hours ago
1
1
@kong that comes from the fact that a polynomial of real coefficients can only have real roots or complex roots in conjugate pairs. That polynomial is the characteristic equation polynomial of the LCCDE equation associated with the real system, which has real coefficients by definition.
– Fat32
1 hour ago
@kong that comes from the fact that a polynomial of real coefficients can only have real roots or complex roots in conjugate pairs. That polynomial is the characteristic equation polynomial of the LCCDE equation associated with the real system, which has real coefficients by definition.
– Fat32
1 hour ago
add a comment |Â
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Please clarify what you mean by “peakâ€Â; do you mean a frequency when the frequency response goes to infinity or is just at a maximum? What about having a peak at DC?
– Dan Boschen
4 hours ago
Also is your thinking restricted to only real inputs (sinusoids) or are you able to understand yet what a complex exponential frequency is? (Since you said this was all new to you)
– Dan Boschen
4 hours ago
@DanBoschen I meant when the frequency goes to infinity. Does having a peak at DC or not change anything ? I know Euler's formula.. is that called a complex exponential frequency ?
– kong
4 hours ago