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This exercise is aimed at showing that zero-padding in the time domain interpolates the frequency domain. Since this is homework, I'll give you the beginning of the solution.

Just like you did, start with:

\begin{align} Y[k] &= \sum_{k=0}^{N-1}x[n]e^{-j2\pi k\frac{n}{2N}}\\ &= \sum_{k=0}^{N-1}x[n]e^{j2\pi k\frac{n}{2N}}e^{-j2\pi k\frac{n}{N}}\\ &= \mathcal{F}\left\{x[n]e^{j2\pi k\frac{n}{2N}}\right\} \end{align} where $\mathcal{F}$ denotes the DFT operator.

Next, we know that multiplication in the time domain is convolution in the frequency domain, so: $$Y[k] = X[k] * \mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\}\tag{1}$$

The next step is to compute $\mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\}$:

\begin{align} \mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\} &= \sum_{k=0}^{N-1}e^{j\pi k\frac{n}{N}}e^{-j2\pi k\frac{n}{N}}\\ &= \sum_{k=0}^{N-1}e^{-j\pi k\frac{n}{N}} \end{align}

Now here is a crucial result for geometric series: $$\sum_{k=0}^{N-1}a^n = \frac{1-a^N}{1-a}$$ So now we have: $$\mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\} = \frac{1-e^{-j\pi k}}{1-e^{-j\pi\frac{k}{N}}}\tag{2}$$ You correctly identified that $e^{-j\pi k} = (-1)^k$, however let's not make that leap. Instead, do you see a way to modify $(2)$ to make $\sin()$ appear on both the numerator and denominator? Use Euler's formula: $$\sin(\theta) = \frac{e^{\theta} - e^{-\theta}}{2j}$$

Ok so as you've correctly calculated, the final expression for $(1)$ is:

$$Y[k] = X[k] * \underbrace{e^{-j\omega_k\big(\frac{N-1}{2}\big)}\frac{\sin(N\omega_k/2)}{\sin(\omega_k/2)}}_{G[\omega_k]}$$ where $\omega_k = \frac{\pi k}{N}$ for simplicity.

$G[\omega_k]$ is an interpolation function. When you convolve this function with $X[k]$, you interpolate $X[k]$. So by zero padding $x[n]$, we interpolated its spectrum $X[k]$.

If you want to take it a step further, here is the convolution sum:

$$Y[l] = \sum_{k=0}^{L-1}X[k]G[\omega_l-\omega_k] = \sum_{k=0}^{N-1}X[k]G[\pi l/2N - \pi k/N]$$

$l = 0, \cdots, 2N-1$

with $L = 2N, \omega_l = \pi l/L, \omega_k = \pi k/N$

This exercise is aimed at showing that zero-padding in the time domain interpolates the frequency domain. Since this is homework, I'll give you the beginning of the solution.

Just like you did, start with:

\begin{align} Y[k] &= \sum_{k=0}^{N-1}x[n]e^{-j2\pi k\frac{n}{2N}}\\ &= \sum_{k=0}^{N-1}x[n]e^{j2\pi k\frac{n}{2N}}e^{-j2\pi k\frac{n}{N}}\\ &= \mathcal{F}\left\{x[n]e^{j2\pi k\frac{n}{2N}}\right\} \end{align} where $\mathcal{F}$ denotes the DFT operator.

Next, we know that multiplication in the time domain is convolution in the frequency domain, so: $$Y[k] = X[k] * \mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\}\tag{1}$$

The next step is to compute $\mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\}$:

\begin{align} \mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\} &= \sum_{k=0}^{N-1}e^{j\pi k\frac{n}{N}}e^{-j2\pi k\frac{n}{N}}\\ &= \sum_{k=0}^{N-1}e^{-j\pi k\frac{n}{N}} \end{align}

Now here is a crucial result for geometric series: $$\sum_{k=0}^{N-1}a^n = \frac{1-a^N}{1-a}$$ So now we have: $$\mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\} = \frac{1-e^{-j\pi k}}{1-e^{-j\pi\frac{k}{N}}}\tag{2}$$

Using Euler’s formula to modify $(2)$, the final expression for $(1)$ is:

$$Y[k] = X[k] * \underbrace{e^{-j\omega_k\big(\frac{N-1}{2}\big)}\frac{\sin(N\omega_k/2)}{\sin(\omega_k/2)}}_{G[\omega_k]}$$ where $\omega_k = \frac{\pi k}{N}$ for simplicity.

$G[\omega_k]$ is an interpolation function. When you convolve this function with $X[k]$, you interpolate $X[k]$. So by zero padding $x[n]$, we interpolated its spectrum $X[k]$.

If you want to take it a step further, here is the convolution sum:

$$Y[l] = \sum_{k=0}^{L-1}X[k]G[\omega_l-\omega_k] = \sum_{k=0}^{N-1}X[k]G[\pi l/2N - \pi k/N]$$

$l = 0, \cdots, 2N-1$

with $L = 2N, \omega_l = \pi l/L, \omega_k = \pi k/N$

This exercise is aimed at showing that zero-padding in the time domain interpolates the frequency domain. Since this is homework, I'll give you the beginning of the solution.

Just like you did, start with:

\begin{align} Y[k] &= \sum_{k=0}^{N-1}x[n]e^{-j2\pi k\frac{n}{2N}}\\ &= \sum_{k=0}^{N-1}x[n]e^{j2\pi k\frac{n}{2N}}e^{-j2\pi k\frac{n}{N}}\\ &= \mathcal{F}\left\{x[n]e^{j2\pi k\frac{n}{2N}}\right\} \end{align} where $\mathcal{F}$ denotes the DFT operator.

Next, we know that multiplication in the time domain is convolution in the frequency domain, so: $$Y[k] = X[k] * \mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\}\tag{1}$$

The next step is to compute $\mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\}$:

\begin{align} \mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\} &= \sum_{k=0}^{N-1}e^{j\pi k\frac{n}{N}}e^{-j2\pi k\frac{n}{N}}\\ &= \sum_{k=0}^{N-1}e^{-j\pi k\frac{n}{N}} \end{align}

Now here is a crucial result for geometric series: $$\sum_{k=0}^{N-1}a^n = \frac{1-a^N}{1-a}$$ So now we have: $$\mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\} = \frac{1-e^{-j\pi k}}{1-e^{-j\pi\frac{k}{N}}}\tag{2}$$ You correctly identified that $e^{-j\pi k} = (-1)^k$, however let's not make that leap. Instead, do you see a way to modify $(2)$ to make $\sin()$ appear on both the numerator and denominator? Use Euler's formula: $$\sin(\theta) = \frac{e^{\theta} - e^{-\theta}}{2j}$$

Ok so as you've correctly calculated, the final expression for $(1)$ is:

$$Y[k] = X[k] * \underbrace{e^{-j\omega_k\big(\frac{N-1}{2}\big)}\frac{\sin(N\omega_k/2)}{\sin(\omega_k/2)}}_{G[\omega_k]}$$ where $\omega_k = \frac{\pi k}{N}$ for simplicity.

$G[\omega_k]$ is an interpolation function. When you convolve this function with $X[k]$, you interpolate $X[k]$. So by zero padding $x[n]$, we interpolated its spectrum $X[k]$.

Hope that helped.

This exercise is aimed at showing that zero-padding in the time domain interpolates the frequency domain. Since this is homework, I'll give you the beginning of the solution.

Just like you did, start with:

\begin{align} Y[k] &= \sum_{k=0}^{N-1}x[n]e^{-j2\pi k\frac{n}{2N}}\\ &= \sum_{k=0}^{N-1}x[n]e^{j2\pi k\frac{n}{2N}}e^{-j2\pi k\frac{n}{N}}\\ &= \mathcal{F}\left\{x[n]e^{j2\pi k\frac{n}{2N}}\right\} \end{align} where $\mathcal{F}$ denotes the DFT operator.

Next, we know that multiplication in the time domain is convolution in the frequency domain, so: $$Y[k] = X[k] * \mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\}\tag{1}$$

The next step is to compute $\mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\}$:

\begin{align} \mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\} &= \sum_{k=0}^{N-1}e^{j\pi k\frac{n}{N}}e^{-j2\pi k\frac{n}{N}}\\ &= \sum_{k=0}^{N-1}e^{-j\pi k\frac{n}{N}} \end{align}

Now here is a crucial result for geometric series: $$\sum_{k=0}^{N-1}a^n = \frac{1-a^N}{1-a}$$ So now we have: $$\mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\} = \frac{1-e^{-j\pi k}}{1-e^{-j\pi\frac{k}{N}}}\tag{2}$$ You correctly identified that $e^{-j\pi k} = (-1)^k$, however let's not make that leap. Instead, do you see a way to modify $(2)$ to make $\sin()$ appear on both the numerator and denominator? Use Euler's formula: $$\sin(\theta) = \frac{e^{\theta} - e^{-\theta}}{2j}$$

Ok so as you've correctly calculated, the final expression for $(1)$ is:

$$Y[k] = X[k] * \underbrace{e^{-j\omega_k\big(\frac{N-1}{2}\big)}\frac{\sin(N\omega_k/2)}{\sin(\omega_k/2)}}_{G[\omega_k]}$$ where $\omega_k = \frac{\pi k}{N}$ for simplicity.

$G[\omega_k]$ is an interpolation function. When you convolve this function with $X[k]$, you interpolate $X[k]$. So by zero padding $x[n]$, we interpolated its spectrum $X[k]$.

If you want to take it a step further, here is the convolution sum:

$$Y[l] = \sum_{k=0}^{L-1}X[k]G[\omega_l-\omega_k] = \sum_{k=0}^{N-1}X[k]G[\pi l/2N - \pi k/N]$$

$l = 0, \cdots, 2N-1$

with $L = 2N, \omega_l = \pi l/L, \omega_k = \pi k/N$

This exercise is aimed at showing that zero-padding in the time domain interpolates the frequency domain. Since this is homework, I'll give you the beginning of the solution.

Just like you did, start with:

\begin{align} Y[k] &= \sum_{k=0}^{N-1}x[n]e^{-j2\pi k\frac{n}{2N}}\\ &= \sum_{k=0}^{N-1}x[n]e^{j2\pi k\frac{n}{2N}}e^{-j2\pi k\frac{n}{N}}\\ &= \mathcal{F}\left\{x[n]e^{j2\pi k\frac{n}{2N}}\right\} \end{align} where $\mathcal{F}$ denotes the DFT operator.

Next, we know that multiplication in the time domain is convolution in the frequency domain, so: $$Y[k] = X[k] * \mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\}$$

The next step is to compute $\mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\}$:

\begin{align} \mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\} &= \sum_{k=0}^{N-1}e^{j\pi k\frac{n}{N}}e^{-j2\pi k\frac{n}{N}}\\ &= \sum_{k=0}^{N-1}e^{-j\pi k\frac{n}{N}} \end{align}

Now here is a crucial result for geometric series: $$\sum_{k=0}^{N-1}a^n = \frac{1-a^N}{1-a}$$ So now we have: $$\mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\} = \frac{1-e^{-j\pi k}}{1-e^{-j\pi\frac{k}{N}}}\tag{1}$$ You correctly identified that $e^{-j\pi k} = (-1)^k$, however let's not make that leap. Instead, do you see a way to modify $(1)$ to make $\sin()$ appear on both the numerator and denominator? Use Euler's formula: $$\sin(\theta) = \frac{e^{\theta} - e^{-\theta}}{2j}$$

This exercise is aimed at showing that zero-padding in the time domain interpolates the frequency domain. Since this is homework, I'll give you the beginning of the solution.

Just like you did, start with:

\begin{align} Y[k] &= \sum_{k=0}^{N-1}x[n]e^{-j2\pi k\frac{n}{2N}}\\ &= \sum_{k=0}^{N-1}x[n]e^{j2\pi k\frac{n}{2N}}e^{-j2\pi k\frac{n}{N}}\\ &= \mathcal{F}\left\{x[n]e^{j2\pi k\frac{n}{2N}}\right\} \end{align} where $\mathcal{F}$ denotes the DFT operator.

Next, we know that multiplication in the time domain is convolution in the frequency domain, so: $$Y[k] = X[k] * \mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\}\tag{1}$$

The next step is to compute $\mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\}$:

\begin{align} \mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\} &= \sum_{k=0}^{N-1}e^{j\pi k\frac{n}{N}}e^{-j2\pi k\frac{n}{N}}\\ &= \sum_{k=0}^{N-1}e^{-j\pi k\frac{n}{N}} \end{align}

Now here is a crucial result for geometric series: $$\sum_{k=0}^{N-1}a^n = \frac{1-a^N}{1-a}$$ So now we have: $$\mathcal{F}\left\{e^{j\pi k\frac{n}{N}}\right\} = \frac{1-e^{-j\pi k}}{1-e^{-j\pi\frac{k}{N}}}\tag{2}$$ You correctly identified that $e^{-j\pi k} = (-1)^k$, however let's not make that leap. Instead, do you see a way to modify $(2)$ to make $\sin()$ appear on both the numerator and denominator? Use Euler's formula: $$\sin(\theta) = \frac{e^{\theta} - e^{-\theta}}{2j}$$

Ok so as you've correctly calculated, the final expression for $(1)$ is:

$$Y[k] = X[k] * \underbrace{e^{-j\omega_k\big(\frac{N-1}{2}\big)}\frac{\sin(N\omega_k/2)}{\sin(\omega_k/2)}}_{G[\omega_k]}$$ where $\omega_k = \frac{\pi k}{N}$ for simplicity.

$G[\omega_k]$ is an interpolation function. When you convolve this function with $X[k]$, you interpolate $X[k]$. So by zero padding $x[n]$, we interpolated its spectrum $X[k]$.

Hope that helped.