BE Computer Engineering (IOE, TU) Digital Signal Analysis and Processing (IOE, EX 701 / ENEX 416) Question Paper 2078 Nepal

(a) Discrete-time LTI system; causality and stability conditions

A discrete-time LTI system is a system that is both linear (obeys superposition: $T\{a x_1[n]+b x_2[n]\}=a\,T\{x_1[n]\}+b\,T\{x_2[n]\}$) and time-invariant (a shift in input produces an identical shift in output). Such a system is completely characterized by its impulse response $h[n]=T\{\delta[n]\}$, and its output is the convolution

$$y[n]=x[n]*h[n]=\sum_{k=-\infty}^{\infty}x[k]\,h[n-k].$$

(i) Causality. A system is causal if the output at time $n$ depends only on present and past inputs. Writing $y[n]=\sum_k h[k]\,x[n-k]$, the term $h[k]x[n-k]$ uses a future input when $k<0$. To exclude all future inputs we require

$$\boxed{h[n]=0 \quad \text{for } n<0.}$$

Proof: If $h[k]=0$ for $k<0$, then $y[n]=\sum_{k\ge 0} h[k]x[n-k]$ involves only $x[n],x[n-1],\dots$ (past/present). Conversely, if $h[m]\neq0$ for some $m<0$, then applying $x[n]=\delta[n]$ gives $y[m]=h[m]\neq0$ at time $m<0$, i.e. output appears before the input — non-causal.

(ii) BIBO stability. A system is BIBO stable if every bounded input produces a bounded output. The necessary and sufficient condition is absolute summability of the impulse response:

$$\boxed{\sum_{n=-\infty}^{\infty}|h[n]|<\infty.}$$

Proof (sufficiency): If $|x[n]|\le M$, then $|y[n]|=\big|\sum_k h[k]x[n-k]\big|\le M\sum_k|h[k]|<\infty$. (Necessity): Choose the bounded input $x[n]=\operatorname{sgn}(h[-n])$; then $y[0]=\sum_k|h[k]|$, which is unbounded if the sum diverges.

(b) Output by linear convolution

Given $h[n]=\left(\tfrac12\right)^n u[n]$ and $x[n]=u[n]-u[n-4]=\{1,1,1,1\}$ for $n=0,1,2,3$ (and 0 elsewhere).

$$y[n]=\sum_{k=0}^{3}x[k]\,h[n-k]=\sum_{k=\max(0,n-3)}^{\min(3,n)}\left(\tfrac12\right)^{n-k},\qquad 0\le n.$$

For $0\le n\le 3$ (rising part):

$$y[n]=\sum_{k=0}^{n}\left(\tfrac12\right)^{n-k}=\sum_{m=0}^{n}\left(\tfrac12\right)^{m}=2-\left(\tfrac12\right)^{n}.$$

So $y[0]=1,\; y[1]=1.5,\; y[2]=1.75,\; y[3]=1.875.$

For $n\ge 4$ (decaying tail): all four input samples ($k=0..3$) contribute:

$$y[n]=\sum_{k=0}^{3}\left(\tfrac12\right)^{n-k}=\left(\tfrac12\right)^{n}\sum_{k=0}^{3}2^{k}=\left(\tfrac12\right)^{n}(15)=15\left(\tfrac12\right)^{n}.$$

Check $n=4$: $15/16=0.9375$.

Result:

$$y[n]=\begin{cases}2-\left(\tfrac12\right)^{n}, & 0\le n\le 3\\[4pt] 15\left(\tfrac12\right)^{n}, & n\ge 4\\ 0,&n<0\end{cases}$$

Numerically: $\{1,\,1.5,\,1.75,\,1.875,\,0.9375,\,0.4688,\dots\}$.

Stability: $\sum_{n=0}^{\infty}\left|\left(\tfrac12\right)^{n}\right|=\dfrac{1}{1-\tfrac12}=2<\infty$, so $h[n]$ is absolutely summable and the system is BIBO stable.

(a) System function $H(z)$ and ROC

Taking the Z-transform of $y[n]-\tfrac34 y[n-1]+\tfrac18 y[n-2]=x[n]+x[n-1]$:

$$Y(z)\Big(1-\tfrac34 z^{-1}+\tfrac18 z^{-2}\Big)=X(z)\big(1+z^{-1}\big).$$ $$\boxed{H(z)=\frac{1+z^{-1}}{1-\tfrac34 z^{-1}+\tfrac18 z^{-2}}=\frac{1+z^{-1}}{\left(1-\tfrac12 z^{-1}\right)\left(1-\tfrac14 z^{-1}\right)}.}$$

Poles at $z=\tfrac12$ and $z=\tfrac14$; zero at $z=-1$ (and a trivial zero at $z=0$). For a causal system the ROC is outside the outermost pole:

$$\text{ROC}:\ |z|>\tfrac12.$$

(b) Pole-zero plot and stability

• Poles ($\times$): $z=0.5$ and $z=0.25$ on the positive real axis, inside the unit circle.
• Zero ($\circ$): $z=-1$ on the unit circle (negative real axis).

Description of the diagram: draw the unit circle; mark two $\times$ on the positive real axis at $0.25$ and $0.5$, and one $\circ$ at $-1$. Since both poles lie inside the unit circle and the causal ROC $|z|>0.5$ includes the unit circle, the system is BIBO stable.

(c) Impulse response by partial fractions

Expand $H(z)$ in $z^{-1}$. Write

$$H(z)=\frac{1+z^{-1}}{\left(1-\tfrac12 z^{-1}\right)\left(1-\tfrac14 z^{-1}\right)}=\frac{A}{1-\tfrac12 z^{-1}}+\frac{B}{1-\tfrac14 z^{-1}}.$$

Multiply out and equate. Let $u=z^{-1}$:

$$A\left(1-\tfrac14 u\right)+B\left(1-\tfrac12 u\right)=1+u.$$

Constant: $A+B=1$. Coefficient of $u$: $-\tfrac14 A-\tfrac12 B=1$. From these: $-\tfrac14 A-\tfrac12(1-A)=1\Rightarrow \tfrac14 A-\tfrac12=1\Rightarrow A=6$, hence $B=-5$.

Using $\dfrac{1}{1-a z^{-1}}\leftrightarrow a^{n}u[n]$ (causal):

$$\boxed{h[n]=6\left(\tfrac12\right)^{n}u[n]-5\left(\tfrac14\right)^{n}u[n].}$$

Check: $h[0]=6-5=1$ ✓ (matches the constant numerator term); $h[1]=6(0.5)-5(0.25)=3-1.25=1.75$.

(d) Magnitude response and filter type

At $\omega=0$ ($z=1$): $H(1)=\dfrac{1+1}{1-0.75+0.125}=\dfrac{2}{0.375}=5.33$ (large gain at DC). At $\omega=\pi$ ($z=-1$): $H(-1)=\dfrac{1-1}{\dots}=0$ (zero at $z=-1$ forces zero gain at $\omega=\pi$). Thus $|H(e^{j\omega})|$ is maximum at $\omega=0$ and falls to zero at $\omega=\pi$ — a monotonically decreasing curve. The system behaves as a low-pass filter.

(a) Impulse invariance vs bilinear transformation

Impulse invariance method. The digital filter is designed so that its impulse response is a sampled version of the analog filter's impulse response: $h[n]=T_s\,h_a(nT_s)$. If $H_a(s)=\sum_k \dfrac{A_k}{s-p_k}$, the mapping of each pole is

$$\frac{A_k}{s-p_k}\ \longrightarrow\ \frac{T_s\,A_k}{1-e^{p_k T_s}z^{-1}},\qquad s=p_k \mapsto z=e^{p_k T_s}.$$

The analog frequency axis maps linearly ($\Omega \to \omega=\Omega T_s$), but because sampling is periodic in frequency, aliasing occurs — it is suitable only for band-limited (low-pass / band-pass) analog prototypes, not high-pass.

Bilinear transformation method. It maps the entire $j\Omega$ axis onto the unit circle once, avoiding aliasing, using

$$\boxed{s=\frac{2}{T}\,\frac{1-z^{-1}}{1+z^{-1}}}\qquad\Leftrightarrow\qquad z=\frac{1+(T/2)s}{1-(T/2)s}.$$

The left-half $s$-plane maps inside the unit circle, so stability is preserved. However the frequency axis is mapped non-linearly:

$$\Omega=\frac{2}{T}\tan\!\left(\frac{\omega}{2}\right)\qquad\text{(frequency warping).}$$

This compresses high analog frequencies into $\omega\le\pi$. Frequency warping distorts the frequency axis, so critical band edges must be pre-warped: $\Omega_c=\dfrac{2}{T}\tan\!\left(\dfrac{\omega_c}{2}\right)$ before applying the transform.

(b) First-order low-pass IIR design (bilinear)

Given $H_a(s)=\dfrac{\Omega_c}{s+\Omega_c}$, digital cut-off $\omega_c=0.2\pi$, $T=1$ s.

Step 1 — Pre-warp the cut-off:

$$\Omega_c=\frac{2}{T}\tan\!\left(\frac{\omega_c}{2}\right)=2\tan(0.1\pi)=2\tan(18^\circ)=2(0.3249)=0.6498\ \text{rad/s}.$$

Step 2 — Apply $s=\dfrac{2}{T}\dfrac{1-z^{-1}}{1+z^{-1}}=2\dfrac{1-z^{-1}}{1+z^{-1}}$:

$$H(z)=\frac{\Omega_c}{\dfrac{2(1-z^{-1})}{1+z^{-1}}+\Omega_c}=\frac{\Omega_c(1+z^{-1})}{2(1-z^{-1})+\Omega_c(1+z^{-1})}.$$

Step 3 — Substitute $\Omega_c=0.6498$: Numerator $=0.6498(1+z^{-1})$. Denominator $=(2+0.6498)+(-2+0.6498)z^{-1}=2.6498-1.3502z^{-1}$.

$$H(z)=\frac{0.6498(1+z^{-1})}{2.6498-1.3502z^{-1}}.$$

Normalize by dividing by $2.6498$:

$$\boxed{H(z)=\frac{0.2452\,(1+z^{-1})}{1-0.5096\,z^{-1}}.}$$

Check (DC gain): at $z=1$, $H(1)=\dfrac{0.2452(2)}{1-0.5096}=\dfrac{0.4904}{0.4904}=1$ — unity gain at DC, confirming a correct low-pass design. The pole at $z=0.5096$ (inside the unit circle) confirms stability.

(a) Decimation-in-time (DIT) radix-2 FFT derivation

The $N$-point DFT is $X[k]=\sum_{n=0}^{N-1}x[n]W_N^{nk}$, where $W_N=e^{-j2\pi/N}$. Split the sum into even ($n=2r$) and odd ($n=2r+1$) indexed samples:

$$X[k]=\sum_{r=0}^{N/2-1}x[2r]W_N^{2rk}+\sum_{r=0}^{N/2-1}x[2r+1]W_N^{(2r+1)k}.$$

Using $W_N^{2rk}=W_{N/2}^{rk}$ and factoring $W_N^{k}$ from the odd sum:

$$\boxed{X[k]=\underbrace{G[k]}_{\text{DFT of even part}}+W_N^{k}\underbrace{H[k]}_{\text{DFT of odd part}},\quad k=0,\dots,N/2-1.}$$

Because $G[k]$ and $H[k]$ are periodic with period $N/2$ and $W_N^{k+N/2}=-W_N^{k}$:

$$X[k+N/2]=G[k]-W_N^{k}H[k].$$

This pair forms the butterfly: two outputs from two inputs using one complex multiply by the twiddle factor $W_N^{k}$. Recursively splitting each $N/2$-DFT gives $\log_2 N$ stages.

$N=8$ signal-flow graph (described). Inputs are applied in bit-reversed order: $x[0],x[4],x[2],x[6],x[1],x[5],x[3],x[7]$. There are $\log_2 8 = 3$ stages, each with $N/2=4$ butterflies (12 butterflies total).

• Stage 1: four 2-point butterflies using twiddle $W_2^0=1$ on adjacent pairs.
• Stage 2: combine into 4-point DFTs using $W_4^0,W_4^1$.
• Stage 3: combine into the 8-point DFT using $W_8^0,W_8^1,W_8^2,W_8^3$. Outputs $X[0]\dots X[7]$ emerge in natural order. Each butterfly: top output $=a+W\,b$, bottom output $=a-W\,b$.

Stage1      Stage2        Stage3
x[0]\ /--•--\        /----•----\  X[0]
x[4]/ \  W2  \      / W4   /     X[1]
x[2]\ /--•--/  W4  /      \ W8   X[2]
x[6]/ \        \  /        \     X[3]
x[1]...                         X[4]
   (bit-reversed input -> natural-order output)

(b) Computational comparison for $N=1024$

Direct DFT: $N^2$ complex multiplications and $N(N-1)$ complex additions.

• Multiplications $=1024^2=1{,}048{,}576$.
• Additions $=1024\times1023=1{,}047{,}552$.

Radix-2 FFT: $\dfrac{N}{2}\log_2 N$ complex multiplications and $N\log_2 N$ complex additions. With $\log_2 1024=10$:

• Multiplications $=\dfrac{1024}{2}\times10=5120$.
• Additions $=1024\times10=10{,}240$.

Savings (multiplications): $\dfrac{1{,}048{,}576}{5120}\approx 205\times$ fewer multiplications. In general the speed-up factor is $\dfrac{N^2}{(N/2)\log_2 N}=\dfrac{2N}{\log_2 N}$, which grows with $N$ — the FFT makes large-$N$ spectral analysis computationally feasible.

4-point DFT of $x[n]=\{1,2,3,4\}$

Using $X[k]=\sum_{n=0}^{3}x[n]\,e^{-j2\pi kn/4}$ with $W_4=e^{-j\pi/2}=-j$, so $W_4^{0}=1,\ W_4^{1}=-j,\ W_4^{2}=-1,\ W_4^{3}=j$.

$X[0]$ (sum): $1+2+3+4=10$.

$X[1]$: $1+2(-j)+3(-1)+4(j)=(1-3)+j(-2+4)=-2+2j$.

$X[2]$: $1+2(-1)+3(1)+4(-1)=1-2+3-4=-2$.

$X[3]$: $1+2(j)+3(-1)+4(-j)=(1-3)+j(2-4)=-2-2j$.

$$\boxed{X[k]=\{\,10,\ -2+2j,\ -2,\ -2-2j\,\}}$$

Magnitude and phase:

Note the conjugate symmetry $X[3]=X[1]^{*}$, as expected for a real input sequence.

Nyquist sampling theorem

A band-limited signal containing no frequency components above $f_{max}$ can be perfectly reconstructed from its samples if it is sampled at a rate $f_s>2f_{max}$. The minimum rate $2f_{max}$ is the Nyquist rate, and $f_s/2$ is the Nyquist (folding) frequency. Frequencies above $f_s/2$ are aliased (folded back) into the band $[0,f_s/2]$.

Analysis of the given signal

$x_a(t)=3\cos(2\pi\cdot1000\,t)+2\cos(2\pi\cdot3000\,t)$, sampled at $f_s=4000$ Hz $\Rightarrow$ folding frequency $=2000$ Hz.

• 1000 Hz component: $1000<2000$ Hz, so it is below the folding frequency — sampled correctly (not aliased). It appears at 1000 Hz.
• 3000 Hz component: $3000>2000$ Hz, so it is aliased. Its apparent (alias) frequency is

$$f_{alias}=|f-f_s|=|3000-4000|=1000\ \text{Hz}.$$

So the 3000 Hz tone masquerades as a 1000 Hz tone.

Reconstructed signal

Both components fall at 1000 Hz after sampling and add (same amplitude/phase form of cosine):

$$\hat{x}(t)=3\cos(2\pi\cdot1000\,t)+2\cos(2\pi\cdot1000\,t)=5\cos(2\pi\cdot1000\,t).$$

The reconstructed signal contains only a single 1000 Hz component of amplitude 5 — the 3000 Hz tone is irretrievably lost to aliasing. To avoid this, $f_s$ must exceed $2\times3000=6000$ Hz.

Window method of FIR filter design

Procedure. Start from the desired (ideal) frequency response $H_d(e^{j\omega})$ and obtain its ideal impulse response $h_d[n]$ by the inverse DTFT (e.g., for an ideal low-pass filter $h_d[n]=\dfrac{\sin(\omega_c n)}{\pi n}$). This is infinite and non-causal, so it is truncated and tapered by multiplying with a finite-length window $w[n]$:

$$h[n]=h_d[n]\,w[n],\qquad 0\le n\le N-1.$$

Multiplication in time = convolution in frequency: $H(e^{j\omega})=H_d * W$. The window's spectrum has a main lobe (which smears the ideal transition, setting the transition-band width) and side lobes (which produce passband/stopband ripple, setting the stopband attenuation). Choosing the window trades transition width against attenuation.

Comparison of windows

Effect on the designed filter

• Rectangular has the narrowest main lobe, giving the sharpest (narrowest) transition band, but its high $-13$ dB side lobes cause large ripple (Gibbs phenomenon) and poor stopband attenuation (~21 dB).
• Hamming lowers side lobes to $-41$ dB, giving ~53 dB stopband attenuation at the cost of a wider transition band (about twice the rectangular).
• Blackman has the lowest side lobes ($-58$ dB) and the best stopband attenuation (~74 dB), but the widest transition band.

Trade-off: lower side-lobe level $\Rightarrow$ greater stopband attenuation but wider main lobe $\Rightarrow$ wider transition band. The window is chosen to meet the required attenuation, and $N$ is increased to sharpen the transition.

Linear convolution vs correlation

In short, correlation is convolution with one sequence time-reversed: $r_{xy}[l]=x[l]*y[-l]$.

Cross-correlation of $x[n]=\{1,2,1\}$ and $y[n]=\{1,-1,2\}$

Using $r_{xy}[l]=\sum_n x[n]\,y[n-l]$ (slide $y$ relative to $x$). Both length-3 sequences (indices 0..2), so $l$ ranges from $-2$ to $+2$.

• $l=-2$: overlap $x[0]y[2]=1\cdot2=2$. $\Rightarrow r_{xy}[-2]=2$.
• $l=-1$: $x[0]y[1]+x[1]y[2]=1(-1)+2(2)=-1+4=3$. $\Rightarrow 3$.
• $l=0$: $x[0]y[0]+x[1]y[1]+x[2]y[2]=1(1)+2(-1)+1(2)=1-2+2=1$. $\Rightarrow 1$.
• $l=+1$: $x[1]y[0]+x[2]y[1]=2(1)+1(-1)=2-1=1$. $\Rightarrow 1$.
• $l=+2$: $x[2]y[0]=1\cdot1=1$. $\Rightarrow 1$.

$$\boxed{r_{xy}[l]=\{\,2,\ 3,\ \underline{1},\ 1,\ 1\,\},\quad l=-2,-1,0,1,2}$$

where the underlined value is at lag $l=0$.

(a) Z-transform and ROC of $x[n]=a^n u[n]+b^n u[-n-1]$

Right-sided term $a^n u[n]$:

$$\sum_{n=0}^{\infty}a^n z^{-n}=\frac{1}{1-a z^{-1}},\qquad \text{ROC: } |z|>|a|.$$

Left-sided term $b^n u[-n-1]$ (nonzero for $n\le -1$):

$$\sum_{n=-\infty}^{-1}b^n z^{-n}=-\frac{1}{1-b z^{-1}},\qquad \text{ROC: } |z|<|b|.$$

Therefore

$$\boxed{X(z)=\frac{1}{1-a z^{-1}}-\frac{1}{1-b z^{-1}},\qquad \text{ROC: } |a|<|z|<|b|.}$$

The ROC is the annulus between the two poles. Condition for existence: the two regions overlap only if $|a|<|b|$. If $|a|\ge|b|$ the ROC is empty and the Z-transform does not converge.

(b) Two important properties of the ROC

1. The ROC is a connected annular region ($r_1<|z|<r_2$) centered at the origin, and it contains no poles (poles lie on its boundary).
2. Sided-ness determines the ROC: for a causal/right-sided sequence the ROC is the exterior of the outermost pole ($|z|>r_{max}$); for an anti-causal/left-sided sequence it is the interior of the innermost pole ($|z|<r_{min}$). (Also: an LTI system is stable iff its ROC includes the unit circle.)

Frequency response of $y[n]=x[n]-x[n-1]$ (first difference)

Impulse response $h[n]=\{1,-1\}$, so

$$H(e^{j\omega})=\sum_n h[n]e^{-j\omega n}=1-e^{-j\omega}.$$

Factor out $e^{-j\omega/2}$:

$$H(e^{j\omega})=e^{-j\omega/2}\big(e^{j\omega/2}-e^{-j\omega/2}\big)=e^{-j\omega/2}\cdot 2j\sin\!\left(\tfrac{\omega}{2}\right).$$

Writing $2j=2e^{j\pi/2}$:

$$\boxed{H(e^{j\omega})=2\sin\!\left(\tfrac{\omega}{2}\right)\,e^{\,j\left(\frac{\pi}{2}-\frac{\omega}{2}\right)}.}$$

Magnitude response:

$$|H(e^{j\omega})|=2\left|\sin\!\left(\tfrac{\omega}{2}\right)\right|.$$

At $\omega=0$: $0$. At $\omega=\pi$: $2$. It rises monotonically from 0 to 2 over $0\le\omega\le\pi$ — a high-pass (differentiator-like) characteristic; the sketch is a half-sine rising from the origin to a peak of 2 at $\omega=\pi$.

Phase response:

$$\angle H(e^{j\omega})=\frac{\pi}{2}-\frac{\omega}{2}.$$

Linear phase? The phase $\dfrac{\pi}{2}-\dfrac{\omega}{2}$ is an affine (linear) function of $\omega$ with constant slope $-\tfrac12$. Hence the system has constant group delay $\tau=-\dfrac{d\angle H}{d\omega}=\tfrac12$ sample and is a (generalized) linear-phase system — expected since $h[n]=\{1,-1\}$ is anti-symmetric (Type-IV linear-phase FIR).

(a) Periodicity of $x[n]=\cos(0.3\pi n)$

A discrete-time sinusoid $\cos(\omega_0 n)$ is periodic iff $\dfrac{\omega_0}{2\pi}=\dfrac{k}{N}$ is rational. Here $\omega_0=0.3\pi$, so

$$\frac{\omega_0}{2\pi}=\frac{0.3\pi}{2\pi}=\frac{0.3}{2}=\frac{3}{20}.$$

This is rational, so the signal is periodic. The fundamental period is the smallest integer $N$ with $\omega_0 N=2\pi k$:

$$N=\frac{2\pi k}{0.3\pi}=\frac{20k}{3}\ \Rightarrow\ k=3,\ \boxed{N=20}.$$

(b) Properties of the system $y[n]=n\,x[n]$

Linear? Yes. For inputs $x_1,x_2$: $T\{a x_1+b x_2\}=n(a x_1[n]+b x_2[n])=a\,n x_1[n]+b\,n x_2[n]=a\,y_1[n]+b\,y_2[n]$. Superposition holds.

Time-invariant? No. Shift the input: response to $x[n-n_0]$ is $n\,x[n-n_0]$, but a shifted output is $y[n-n_0]=(n-n_0)x[n-n_0]$. Since $n\,x[n-n_0]\neq(n-n_0)x[n-n_0]$, the system is time-varying (the gain factor $n$ depends explicitly on time).

Stable? No. The multiplier $n$ grows without bound. For the bounded input $x[n]=1$ (all $n\ge0$), $y[n]=n\to\infty$, so a bounded input produces an unbounded output — BIBO unstable.

Summary: $x[n]=\cos(0.3\pi n)$ is periodic with $N=20$; the system $y[n]=n x[n]$ is linear but time-varying and unstable.

DFT–DTFT relationship

For a length-$N$ sequence $x[n]$, the DTFT is the continuous-frequency spectrum

$$X(e^{j\omega})=\sum_{n=0}^{N-1}x[n]e^{-j\omega n}.$$

The $N$-point DFT is obtained by sampling the DTFT uniformly at $N$ equally-spaced frequencies $\omega_k=\dfrac{2\pi k}{N}$:

$$\boxed{X[k]=X(e^{j\omega})\Big|_{\omega=2\pi k/N}=\sum_{n=0}^{N-1}x[n]e^{-j2\pi kn/N},\quad k=0,\dots,N-1.}$$

Thus the DFT is the frequency-sampled DTFT over one period $[0,2\pi)$; equivalently, it corresponds to the DTFT of the periodically-extended sequence. (It also equals the Z-transform evaluated at the $N$ equally-spaced points $z=e^{j2\pi k/N}$ on the unit circle.)

Two important DFT properties

1. Circular shift. A circular time shift by $m$ introduces a linear phase:

$$x\big[(n-m)\bmod N\big]\ \xleftrightarrow{\ \text{DFT}\ }\ X[k]\,e^{-j2\pi km/N}.$$

2. Parseval's theorem. Energy is conserved between time and frequency domains:

$$\sum_{n=0}^{N-1}|x[n]|^{2}=\frac{1}{N}\sum_{k=0}^{N-1}|X[k]|^{2}.$$

(Other valid properties: linearity, circular convolution $x[n]\circledast h[n]\leftrightarrow X[k]H[k]$, and symmetry $X[N-k]=X^{*}[k]$ for real $x[n]$.)