Question about summation in power spectral density formula

In summary, the author of the textbook, Leon Couch, found the PSD (power spectral density) of a digital NRZ pulse train by using the definition of PSD for a periodic signal and showing that the first term of the Fourier series can be expressed as a coefficient of ##A^2/4##, leading to the final equation for PSD. The author's reasoning may seem confusing at first, but it is correct as the sum converges and the desired coefficient of ##A^2/2## is obtained when rearranging the first term.
  • #1
FrankJ777
140
6

Homework Statement


This isn't really a problem I've been given, but questions i have about how the author of my textbook, Leon Couch, Digital and Analog communications Systems, found the PSD (power spectral density) of an digital NRZ pulse train.

Homework Equations


The PSD of a periodic signal is defined as: [itex]
\mathit{P(s)} = \frac{\left |F(f) \right |^{2}}{T}\sum_{k=- \infty}^{\infty}R(k)e^{j2\pi kfT}
[/itex]
The text shows that: [itex] R(0) = \frac{A^{2}}{2}, R(k\neq 0)=\frac{A^{2}}{4} [/itex]
In the next step he shows that: [itex] \mathit{P(s)} = \frac{A^{2}}{4}\frac{\left |F(f) \right |^{2}}{T}\left [1+\sum_{k=- \infty}^{\infty}e^{j2\pi kfT} \right ] [/itex]
This is where I start being confused. It seems as though he factored the R(0) term out of the sum and then factored the [itex] R(0) = \frac{A^{2}}{2}, R(k\neq 0)=\frac{A^{2}}{4} [/itex] factors out of the summation leaving [itex] \mathit{P(s)} = \frac{A^{2}}{4}\frac{\left |F(f) \right |^{2}}{T}\left [1+\sum_{k=- \infty}^{\infty}e^{j2\pi kfT} \right ] [/itex]
To me this doesn't seem right because when factoring the R(0) term out of the summation, the interval of summation is still from negative infinity to infinity, so zero is still in that interval. Also it seems that when he factors all the R(k) factors out of the summation, he doesn't account for the fact that [itex] R(0) = \frac{A^{2}}{2} [/itex]

The Attempt at a Solution


To me it seems that it the step should have looked like this: [itex] \mathit{P(s)} = \frac{A^{2}}{4}\frac{\left |F(f) \right |^{2}}{T}\left [2+\sum_{k=1}^{\infty}e^{j2\pi kfT} +\sum_{k=-\infty}^{-1}e^{j2\pi kfT} \right ] [/itex]
This way R(0) is not in the summation interval after its been pulled out, and the two separate summations account for the intervals from negative infinity to -1, then from 1 to infinity, and R(0) remains separate.
Also this way both when [itex] R(0) = \frac{A^{2}}{2}, R(k\neq 0)=\frac{A^{2}}{4} [/itex] are factored out, the number 2 is left where the R(0) factor was, which i think is correct.
Can someone please point out if my line of thinking is correct, or please explain to me why the text is correct?
Thanks a lot.
[itex] [/itex]
 
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  • #2
FrankJ777 said:
The PSD of a periodic signal is defined as: [itex]
\mathit{P(s)} = \frac{\left |F(f) \right |^{2}}{T}\sum_{k=- \infty}^{\infty}R(k)e^{j2\pi kfT}
[/itex]
The text shows that: [itex] R(0) = \frac{A^{2}}{2}, R(k\neq 0)=\frac{A^{2}}{4} [/itex]
In the next step he shows that: [itex] \mathit{P(s)} = \frac{A^{2}}{4}\frac{\left |F(f) \right |^{2}}{T}\left [1+\sum_{k=- \infty}^{\infty}e^{j2\pi kfT} \right ] [/itex]

As a Fourier series, the first term is ##\frac{A^2/2}{2}=\frac{A^2}{4}## (i.e., divided by ##2##). That, I believe, is what's going on here.

EDIT (Again D: )
OK! My bad.

What's happening here is this. The sum converges, so we rearrange a finite number of terms (i.e., one term) and notice that at ##k=0## we obtain your desired coefficient since ##1+e^{0}=2## and ##2A^2/4=A^2/2##.
 
Last edited:

Related to Question about summation in power spectral density formula

What is the formula for calculating power spectral density?

The formula for power spectral density is P(f) = |FFT(x)|^2 / N, where P(f) represents the power at a specific frequency, FFT(x) is the Fourier transform of the signal x, and N is the total number of samples in the signal.

What is the significance of power spectral density in signal processing?

Power spectral density is a useful tool in signal processing as it allows us to analyze the frequency content of a signal. It provides information about the distribution of power across different frequencies, which can be helpful in identifying dominant frequencies and detecting any anomalies or patterns in the signal.

How is power spectral density related to the Fourier transform?

Power spectral density is essentially the squared magnitude of the Fourier transform of a signal. The Fourier transform converts a signal from the time domain to the frequency domain, and the power spectral density formula is derived from this transform.

Can power spectral density be used for both continuous and discrete signals?

Yes, power spectral density can be used for both continuous and discrete signals. However, for continuous signals, the formula is slightly different and involves taking the limit as the number of samples approaches infinity.

What are some applications of power spectral density?

Power spectral density has various applications in fields such as signal processing, communications, and control systems. It is used for tasks such as filtering, spectrum analysis, and detecting signal characteristics. Additionally, it is also used in fields such as astronomy, geophysics, and biomedical engineering for analyzing signals and detecting patterns or anomalies.

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