Let the linear convolution of two discrete-time finite-length sequences $x[n]$ and $h[n]$, each of length 16, be denoted by $y[n]$. Let $z[n]$ denote the 16-point inverse discrete Fourier transform (IDFT) of the product of the 16-point discrete Fourier transforms (DFTs) of $x[n]$ and $h[n]$. The value of $k$ for which $z[k]=y[k]$ is (A) 0 (B) 1 (C) 15 (D) 16 Solution Step 1: Length of Linear Convolution Both sequences have length 16. Length of linear convolution: $ 16+16-1=31 $ Hence the linear convolution contains the samples $ y[0],y[1],\ldots,y[30] $ Notice that there is no sample $y[31]$. Step 2: What does the IDFT of the Product of DFTs Give? A standard DFT property states: $ \text{IDFT}\{X[k]H[k]\} = \text{Circular Convolution} $ Therefore, $ z[n] = \text{16-point circular convolution of }x[n]\text{ and }h[n]. $ So, $y[n]$ = Linear Convolution $z[n]$ = 16-point Circular Convolution Step 3: Folding (Wrap-A...

Linear vs Circular Convolution Linear convolution vs circular convolution Linear convolution (what nature does) If: x[n] = transmitted signal h[n] = channel impulse response (multipath) Then the received signal is: y[n] = x[n] * h[n] This means: Past symbols spill into future symbols Causes inter-symbol interference (ISI) DFT of y[n] is not a simple product Circular convolution (what DFT assumes) The DFT assumes periodic signals : y[n] = x[n] ⊛ h[n], where indices wrap around modulo N Only under circular convolution does the DFT satisfy: Y[k] = X[k] · H[k] This is what enables one-tap frequency-domain equalization . What happens in the real channel (linear convolution) The channel performs linear convolution: y[n] = Σ l=0 Lh−1 h[l] x[n-l] But because: CP absorbs the “spillover” Receiver discards CP before DFT The useful part of the received symbol becomes: y[n] = Σ l=0 Lh−1 h[l] x[(n-l) mod N] Modulo inde...

When two signals are convolved in the time domain, what frequency components will be present in the frequency domain? Is it similar to (frequency 1 + frequency 2) \text{(frequency 1 + frequency 2)} (frequency 1 + frequency 2) and (frequency 1 - frequency 2) \text{(frequency 1 - frequency 2)} (frequency 1 - frequency 2) ? No, it isn’t. That formula applies to the time-domain multiplication of two sinusoidal signals. According to the Discrete Convolution Theorem, convolution of two discrete signals in the time domain is equivalent to multiplication of their DFTs in the frequency domain: F{x[n] ∗ h[n]} = X[k] ⋅ H[k] where  X[k]  and  H[k]  are the DFTs of  x[n]  and  h[n] , respectively. Thus, the convolution  y[n]  in the time domain can be computed by taking the inverse DFT of the product: y[n] = F -1 {X[k] ⋅ H[k]} In general, the frequency components present in X...

  Suppose, you have two arrays A = [1,2,3,4], and B = [10,11,12]. Then the convolution result, C will be [10,31,64,97,80,80]. The length of C is length of A + length of B -1 . Convolution is a very useful concept in signal processing. We've often taken consideration that transmitted signal is convolved with channel impulse response (CIR) in the time domain. And in the frequency domain they are in multiplication form. Lets discuss how mathematically find the convolution of two arrays or signals. For the above example, the formula for convolution is: \[ C[n] = (A * B)[n] = \sum_{m= 0}^{m} A[m] B[n - m] \]  Where, m is the length of A      and, n varies from 0 to m + p -1   where, p is the length of B   Convolution Formula The convolution of two continuous-time signals \( x(t) \) and \( h(t) \) is given by: \[ y(t) = (x * h)(t) = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) \, d\tau \] ...

In this MATLAB code it is described how to calculate 'BER vs SNR' and 'SER vs SNR' from the Channel Impulse Response (CIR) .  SER =  Symbol error rate For a typical communication system, if we transmit the signal x[n], then received signal will be, y[n] = h[n]*x[n]  + noise Where, h[n] is channel impulse response and '*' indicates convolution operation MATLAB Code clc; clear all; close all; % Channel Impulse Response (CIR) channel_response = [0.85]; % Example channel impulse response coefficients %channel_response = [0.8, 0.1]; % Modulation parameters modulation_order = 4; % QPSK modulation num_symbols = 1000; % Number of symbols to transmit num_bits = 1000*log2(modulation_order); % Signal generation tx_symbols = randi([0, modulation_order-1], 1, num_symbols); tx_signal = qammod(tx_symbols, modulation_order); % Channel simulation rx_signal = conv(tx_signal, channel_response); % Add AWGN (Additive White Gaussian Noise) snr_dB = 0:2:20; ber = zeros(size(snr_dB))...

  When a signal is convolved with an impulse response, the result is the output signal of a linear time-invariant (LTI) system . This process is fundamental in signal processing and is often used to analyze and understand the behavior of systems in response to different inputs. By convolving the input signal with the system's impulse response, we obtain the output signal, providing insights into the system's characteristics, such as filtering, time delays, and frequency response. It provides information about how a communication channel responds to an impulse signal. Here are key points about the Channel Impulse Response: 1. Time Domain Representation:    In the time domain, the channel impulse response represents how the channel reacts to the impulse signal over time. 2. Frequency Domain Representation:    The Channel Impulse Response is also related to the frequency response of the channel. The Fourier Transform of the impulse response gives the frequency resp...

  LTI Systems LTI Systems and Convolution An LTI system is a system that is both linear (obeys the superposition principle) and time-invariant (its behavior doesn't change over time). The output y(t) of an LTI system is determined by the convolution of the input signal x(t) with the system's unique impulse response h(t) . Convolution is a mathematical operation expressed as: y(t) = ∫ -∞ ∞ x(τ) h(t - τ) dτ 1. For a Discrete-Time System In wireless communication, the signal reaches the receiver thru different multi-paths. And they are nothing but time-delayed versions of the same transmitted signal. You always find a relationship between received and transmitted signals is y = h * x + n  which denotes that the transmitted signal 'x' is convolved with the channel coefficients ('h') of the particular channel. Because here time-shifted version of the initially transmitted signal is overlapped with the ...