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DTFT conjugation and frequency shifting

Short answer

Conjugation: x*[n] ↔ X*(e^{−jω}) (sum conjugate). Frequency shifting: e^{jω₀n}x[n] ↔ X(e^{j(ω−ω₀)}) (exponential factor). Both properties follow directly from the DTFT definition.

State and prove conjugate and frequency shifting properties of discrete time Fourier transform.

Let x[n]↔DTFTX(ejω)=∑nx[n]e−jωnx[n] \xleftrightarrow{DTFT} X(e^{j\omega}) = \sum_n x[n]e^{-j\omega n}.

Conjugation property

Statement:

x∗[n]↔DTFTX∗(e−jω)x^*[n] \xleftrightarrow{DTFT} X^*(e^{-j\omega})

Proof:

∑nx∗[n]e−jωn=[∑nx[n]ejωn]∗=[∑nx[n]e−j(−ω)n]∗=X∗(e−jω)\sum_{n}x^*[n]e^{-j\omega n} = \left[\sum_{n}x[n]e^{j\omega n}\right]^* = \left[\sum_{n}x[n]e^{-j(-\omega)n}\right]^* = X^*(e^{-j\omega})

Consequence: if x[n]x[n] is real, X(ejω)=X∗(e−jω)X(e^{j\omega}) = X^*(e^{-j\omega}). The magnitude is then even and the phase is odd.

Frequency shifting property

Statement:

ejω0nx[n]↔DTFTX(ej(ω−ω0))e^{j\omega_0 n}x[n] \xleftrightarrow{DTFT} X(e^{j(\omega - \omega_0)})

Proof:

∑nejω0nx[n]e−jωn=∑nx[n]e−j(ω−ω0)n=X(ej(ω−ω0))\sum_{n}e^{j\omega_0 n}x[n]e^{-j\omega n} = \sum_{n}x[n]e^{-j(\omega - \omega_0)n} = X(e^{j(\omega - \omega_0)})

Multiplying by ejω0ne^{j\omega_0 n} shifts the whole periodic spectrum by ω0\omega_0. Example: (−1)nx[n]=ejπnx[n]↔X(ej(ω−π))(-1)^n x[n] = e^{j\pi n}x[n] \leftrightarrow X(e^{j(\omega - \pi)}), which turns a low-pass spectrum into a high-pass one.

✓ Answer checked by Claude Opus 5.5 (high effort)

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