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DTFT properties; convolution property

Short answer

Convolution in time corresponds to multiplication in frequency: y[n] = x[n]*h[n] <-> Y(e^{jw}) = X(e^{jw})H(e^{jw}). Proof: Y(e^{jw}) = sum_n (sum_k x[k]h[n-k]) e^{-jw n} = sum_k x[k] e^{-jw k} H(e^{jw}) = X(e^{jw})H(e^{jw}).

List out (explain) the properties of Discrete Time Fourier Transform. State and prove the convolution property of Discrete Time Fourier Transform.

Properties of the DTFT

Let x[n]↔X(ejω)x[n] \leftrightarrow X(e^{j\omega}) and y[n]↔Y(ejω)y[n] \leftrightarrow Y(e^{j\omega}).

PropertyTime domainFrequency domain
Periodicityx[n]x[n]X(ej(ω+2π))=X(ejω)X(e^{j(\omega+2\pi)}) = X(e^{j\omega})
Linearityax[n]+by[n]ax[n] + by[n]aX(ejω)+bY(ejω)aX(e^{j\omega}) + bY(e^{j\omega})
Time shiftx[n−n0]x[n - n_0]e−jωn0X(ejω)e^{-j\omega n_0}X(e^{j\omega})
Frequency shiftejω0nx[n]e^{j\omega_0 n}x[n]X(ej(ω−ω0))X(e^{j(\omega-\omega_0)})
Conjugationx∗[n]x^*[n]X∗(e−jω)X^*(e^{-j\omega})
Time reversalx[−n]x[-n]X(e−jω)X(e^{-j\omega})
Differencingx[n]−x[n−1]x[n] - x[n-1](1−e−jω)X(ejω)(1 - e^{-j\omega})X(e^{j\omega})
Differentiation in freq.nx[n]nx[n]jdX(ejω)dωj\frac{dX(e^{j\omega})}{d\omega}
Convolutionx[n]∗y[n]x[n]*y[n]X(ejω)Y(ejω)X(e^{j\omega})Y(e^{j\omega})
Multiplicationx[n]y[n]x[n]y[n]12π∫2πX(ejθ)Y(ej(ω−θ))dθ\frac{1}{2\pi}\int_{2\pi}X(e^{j\theta})Y(e^{j(\omega-\theta)})d\theta
Parseval∑∣x[n]∣2\sum\lvert x[n]\rvert^212π∫2π∣X(ejω)∣2dω\frac{1}{2\pi}\int_{2\pi}\lvert X(e^{j\omega})\rvert^2d\omega

Also, for real x[n]x[n] the DTFT is conjugate symmetric: X(e−jω)=X∗(ejω)X(e^{-j\omega}) = X^*(e^{j\omega}), so the magnitude is even and the phase is odd.

Convolution property: statement

y[n]=x[n]∗h[n]=∑k=−∞∞x[k]h[n−k]  ↔DTFT  Y(ejω)=X(ejω)H(ejω)y[n] = x[n]*h[n] = \sum_{k=-\infty}^{\infty}x[k]h[n-k] \;\xleftrightarrow{DTFT}\; Y(e^{j\omega}) = X(e^{j\omega})H(e^{j\omega})

Proof

Y(ejω)=∑n=−∞∞∑k=−∞∞x[k]h[n−k]e−jωn=∑kx[k]∑nh[n−k]e−jωnY(e^{j\omega}) = \sum_{n=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}x[k]h[n-k]e^{-j\omega n} = \sum_{k}x[k]\sum_{n}h[n-k]e^{-j\omega n}

Put m=n−km = n - k in the inner sum:

∑m=−∞∞h[m]e−jω(m+k)=e−jωkH(ejω)\sum_{m=-\infty}^{\infty}h[m]e^{-j\omega(m+k)} = e^{-j\omega k}H(e^{j\omega})

So

Y(ejω)=H(ejω)∑kx[k]e−jωk=X(ejω)H(ejω)Y(e^{j\omega}) = H(e^{j\omega})\sum_{k}x[k]e^{-j\omega k} = X(e^{j\omega})H(e^{j\omega})

Use: an LTI system's output spectrum is its input spectrum multiplied by the frequency response H(ejω)H(e^{j\omega}).

✓ Answer checked by Claude Opus 5.5 (high effort)

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