The Correct Answer and Explanation is:
To prove Parseval’s theorem, we follow the three parts in the problem statement:
Correct answer:
- Given x[n]∈l2x[n] \in l^2, define y[n]=∑mx[m]x[m−n]y[n] = \sum_m x[m]x[m – n]. This expression is equivalent to convolution:
y[n]=x[n]∗x[−n]y[n] = x[n] * x[-n]
- Using the Fourier transform properties:
F{x[−n]}=X∗(ejω)\mathcal{F}\{x[-n]\} = X^*(e^{j\omega})
then,
Y(ejω)=X(ejω)⋅X∗(ejω)=∣X(ejω)∣2Y(e^{j\omega}) = X(e^{j\omega}) \cdot X^*(e^{j\omega}) = |X(e^{j\omega})|^2
- The energy of the signal in time and frequency domains is:
y[0]=∑mx[m]x[m]=∑m∣x[m]∣2y[0] = \sum_m x[m]x[m] = \sum_m |x[m]|^2
Therefore,
12π∫02π∣X(ejω)∣2dω=∑n∣x[n]∣2\frac{1}{2\pi} \int_0^{2\pi} |X(e^{j\omega})|^2 d\omega = \sum_n |x[n]|^2
Explanation
Parseval’s theorem states that the total energy of a signal in the time domain is equal to the total energy in the frequency domain. This result provides a powerful link between these two representations. To prove it, we begin by defining the autocorrelation function y[n]=∑mx[m]x[m−n]y[n] = \sum_m x[m]x[m – n]. This function measures how the signal correlates with shifted versions of itself, and it can be rewritten as a convolution operation x[n]∗x[−n]x[n] * x[-n] by recognizing the shift and summation pattern.
Next, we apply the Fourier transform to both sides. The convolution in the time domain becomes multiplication in the frequency domain. The key identity is F{x[−n]}=X∗(ejω)\mathcal{F}\{x[-n]\} = X^*(e^{j\omega}), the complex conjugate of the Fourier transform of the signal. Therefore, the transform of the autocorrelation function becomes Y(ejω)=∣X(ejω)∣2Y(e^{j\omega}) = |X(e^{j\omega})|^2.
Finally, evaluating y[0]y[0], which corresponds to the zero shift, gives the total signal energy ∑n∣x[n]∣2\sum_n |x[n]|^2. On the frequency side, the inverse Fourier transform of Y(ejω)Y(e^{j\omega}) evaluated at zero is exactly 12π∫02π∣X(ejω)∣2dω\frac{1}{2\pi} \int_0^{2\pi} |X(e^{j\omega})|^2 d\omega, completing the proof.
This result is especially useful in signal processing because it allows energy computations to shift to whichever domain is more convenient computationally.
