Discrete Fourier Transform by hand I have an assignment where I'm given the DFT of a sequence $x[n]$ as $X[k]=\{4,3,2,1,0,1,2,3\}$ and also
$$y[n] = \left\{ \begin{array}[cc]
xx[n/2] &amp; \text{if n is even} \\
0 &amp; \text{otherwise}
\end{array} \right. $$

and I'm supposed to find and sketch the DFT of $y[n]$.

So $y[n] = \{x[0], 0, x[1], 0 ... x[7], 0\}$ and it's not complicated to find $Y[k]$ if we know $x[n]$

I know how to use the definition of the DFT and IDFT to calculate $x[n]$ but it's a tedious task to do by hand, especially when the sequence is longer than a few items. Is there a quicker way to calculate the DFT and IDFT by hand without using a program like Matlab?
If you use the DFT formula, you get:
$$
Y[k] = \sum_{n=0}^{2N-1}y[n]e^{\frac{-2\pi k n}{2N}} 
$$
Now, substituting the definition of $y[n]$ you get:
$$
Y[k] = \sum_{n=0}^{N-1}x[n]e^{\frac{-2\pi k (2n)}{2N}} = \sum_{n=0}^{N-1}x[n]e^{\frac{-2\pi k n}{N}}
$$
So, for $0\leq k &lt; N$ you get that
$$ Y[k] = X[k]$$
and for $k\geq N$ you get
$$
Y[k] = \sum_{n=0}^{N-1}x[n]e^{\frac{-2\pi k n}{N}} = \sum_{n=0}^{N-1}x[n]e^{\frac{-2\pi (k-N) n}{N}} = Y[k-N]
$$
Therfore, 
$$
Y[k] = \begin{cases}X[k] &amp; 0\leq k &lt; N \\ X[k-N] &amp; k\geq N\end{cases}
$$
or $X[k]=\{4,3,2,1,0,1,2,3,4,3,2,1,0,1,2,3\}$