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What differences does it make?

In one line - The phase information of the Fourier transform changes.

Explanation:

Adding zeros at the start of the signal translates to adding delay.

Consider the signal is $x[n]$. Let $L$ be the original length of the signal. If $N$ zeros are appended at the start of the signal then the resulting signal $y[n]$ can be represented as below: $$y[n] = 0\ for\ 0<=n<N $$ $$y[n] = x[n - N]\ for\ N <= n < (N + L) $$

Consider that $X[k]$ is the DFT (for length $L' = L + N$) of $x[n]$, then the DFT of $y[n]$ will be $X[k] * e^{\frac{j2 \pi Nk}{L'}}$. As can be seen, the additional factor $e^{\frac{j2 \pi Nk}{L'}}$ changes the phase information of the DFT and the phase change is related to the delay / number of zeros added at the start $N$.

What differences does it make?

In one line - The phase information of the Fourier transform changes.

Explanation:

Adding zeros at the start of the signal translates to adding delay.

Consider the signal $x[n]$. Let $L$ be the original length of the signal. If $N$ zeros are appended at the start of the signal then the resulting signal $y[n]$ can be represented as below: $$y[n] = 0\ for\ 0<=n<N $$ $$y[n] = x[n - N]\ for\ N <= n < (N + L) $$

Consider that $X[k]$ is the DFT (for length $L' = L + N$) of $x[n]$, then the DFT of $y[n]$ will be $X[k] * e^{\frac{j2 \pi Nk}{L'}}$. As can be seen, the additional factor $e^{\frac{j2 \pi Nk}{L'}}$ changes the phase information of the DFT and the phase change is related to the delay / number of zeros added at the start $N$.

What differences does it make? Blockquote

In one line - The phase information of the Fourier transform changes.

Explanation:

Adding zeros at the start of the signal translates to adding delay.

Consider the signal is $x[n]$. Let $L$ be the original length of the signal. If $N$ zeros are appended at the start of the signal then the resulting signal $y[n]$ can be represented as below: $$y[n] = 0\ for\ 0<=n<N $$ $$y[n] = x[n - N]\ for\ N <= n < (N + L) $$

Consider that $X[k]$ is the DFT (for length $L' = L + N$) of $x[n]$, then the DFT of $y[n]$ will be $X[k] * e^{\frac{j2 \pi Nk}{L'}}$. As can be seen, the additional factor $e^{\frac{j2 \pi Nk}{L'}}$ changes the phase information of the DFT and the phase change is related to the delay / number of zeros added at the start $N$.

What differences does it make?

In one line - The phase information of the Fourier transform changes.

Explanation:

Adding zeros at the start of the signal translates to adding delay.

Consider the signal is $x[n]$. Let $L$ be the original length of the signal. If $N$ zeros are appended at the start of the signal then the resulting signal $y[n]$ can be represented as below: $$y[n] = 0\ for\ 0<=n<N $$ $$y[n] = x[n - N]\ for\ N <= n < (N + L) $$

Consider that $X[k]$ is the DFT (for length $L' = L + N$) of $x[n]$, then the DFT of $y[n]$ will be $X[k] * e^{\frac{j2 \pi Nk}{L'}}$. As can be seen, the additional factor $e^{\frac{j2 \pi Nk}{L'}}$ changes the phase information of the DFT and the phase change is related to the delay / number of zeros added at the start $N$.

What differences does it make? Blockquote

In one line - The phase information of the Fourier transform changes.

Explanation:

Adding zeros at the start of the signal translates to adding delay.

Consider the signal is $x[n]$. Let $L$ be the original length of the signal. If $N$ zeros are appended at the start of the signal then the resulting signal $y[n]$ can be represented as below: $$y[n] = 0\ for\ 0<=n<N $$ $$y[n] = x[n - N]\ for\ N <= n < (N + L) $$

Consider that $X[k]$ is the DFT (for length $L' = L + N$) of $x[n]$, then the DFT of $x[n - N]$ will be $X[k] * e^{\frac{j2 \pi Nk}{L'}}$. As can be seen, the additional factor $e^{\frac{j2 \pi Nk}{L'}}$ changes the phase information of the DFT and the phase change is related to the delay / number of zeros added at the start $N$.

What differences does it make? Blockquote

In one line - The phase information of the Fourier transform changes.

Explanation:

Adding zeros at the start of the signal translates to adding delay.

Consider the signal is $x[n]$. Let $L$ be the original length of the signal. If $N$ zeros are appended at the start of the signal then the resulting signal $y[n]$ can be represented as below: $$y[n] = 0\ for\ 0<=n<N $$ $$y[n] = x[n - N]\ for\ N <= n < (N + L) $$

Consider that $X[k]$ is the DFT (for length $L' = L + N$) of $x[n]$, then the DFT of $y[n]$ will be $X[k] * e^{\frac{j2 \pi Nk}{L'}}$. As can be seen, the additional factor $e^{\frac{j2 \pi Nk}{L'}}$ changes the phase information of the DFT and the phase change is related to the delay / number of zeros added at the start $N$.