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Parseval's Theorem

Module by: Don Johnson

Summary: More about Parseval's Theorem.

The properties of the discrete-time Fourier transform mirror those of the analog Fourier transform. The DTFT properties table shows similarities and differences. One important common property is Parseval's Theorem.

∑n=-∞∞|sn|2=∫-1212|Sⅇⅈ2πf|2df n s n 2 f 1 2 1 2 S 2 f 2 (1)
To show this important property, we simply substitute the Fourier transform expression into the frequency-domain expression for power.
∫-1212|Sⅇⅈ2πf|2df=∫-1212∑nsnⅇ-ⅈ2πfn∑msn¯ⅇ+ⅈ2πfmdf=∑ n , m snsm¯∫-1212ⅇ+ⅈ2πfm-ndf f 1 2 1 2 S 2 f 2 f 1 2 1 2 n n s n 2 f n m m s n 2 f m n , m n , m s n s m f 1 2 1 2 2 f m n (2)
Using the orthogonality relation, the integral equals δm-n δ m n , where δn δ n is the unit sample. Thus, the double sum collapses into a single sum because nonzero values occur only when n=m n m , giving Parseval's Theorem as a result. We term ∑ns2n n n s n 2 the energy in the discrete-time signal sn s n in spite of the fact that discrete-time signals don't consume (or produce for that matter) energy. This terminology is a carry-over from the analog world.

Exercise 1

Suppose we obtained our discrete-time signal from values of the product st p T s t s t p T s t , where the duration of the component pulses in p T s t p T s t is Δ Δ . How is the discrete-time signal energy related to the total energy contained in st s t ? Assume the signal is bandlimited and that the sampling rate was chosen appropriate to the Sampling Theorem's conditions.

Solution 1

If the sampling frequency exceeds the Nyquist frequency, the spectrum of the samples equals the analog spectrum, but over the normalized analog frequency fT f T . Thus, the energy in the sampled signal equals the original signal's energy multiplied by TT.

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