For T= 0,  Delay = (2+1) = 3ns For T= 1 Delay = (2+1) + (2+1) = 6ns For worst case it is 6 ns Answer (c) 6 ns

Flat vs Frequency-Selective Fading Simulator Flat vs Frequency-Selective Fading Interactive Multipath Channel Simulator Change the channel mode, delay spread, signal bandwidth, Doppler frequency, and SNR. Observe how the channel response changes. Parameters Channel Mode Flat Fading Frequency-Selective Fading Signal Type Pure Sine Wave Wideband Multitone Signal Center Frequency 1.00 MHz Signal Bandwidth 1.00 MHz Maximum Doppler 30 Hz SNR 25 dB Reflected Path Power -3 dB Maximum Path Delay 2.0 μs Number of Paths 4 Run Simulation Channel -- Signal BW -- Coherence BW -- RMS Delay Spread -- 1. Transmitted Signal 2. Channel Impulse Response 3. Channel Frequency Response 4. Received Signal 5. Signal Spectrum 6. Received Envelope 7. Doppler ...

  1. At node X The source current i s ( t ) i_s(t) splits into capacitor current and resistor current: i s ( t ) = i C ( t ) + i R ( t ) i_s(t)=i_C(t)+i_R(t) For the capacitor, i C = C d v C d t = d v C d t i_C=C\frac{dv_C}{dt} =\frac{dv_C}{dt} because C = 1 F C=1F . For the 1 Ω 1\Omega resistor, i R = v C 1 = v C i_R=\frac{v_C}{1}=v_C Therefore, i s = d v C d t + v C i_s=\frac{dv_C}{dt}+v_C so v ˙ C = i s − v C \boxed{\dot v_C=i_s-v_C} This gives the second state equation. 2. For the inductor branch The inductor 1 H 1H and resistor 1 Ω 1\Omega are parallel . Therefore, they have the same voltage . For the inductor, v L = L d i L d t v_L=L\frac{di_L}{dt} Since L = 1 H L=1H , v L = d i L d t v_L=\frac{di_L}{dt} For the parallel 1 Ω 1\Omega resistor, v R = i R v_R=i_R Since their voltages are equal, d i L d t = i R \frac{di_L}{dt}=i_R Now the source current splits between the inductor and resistor: i s = i L + i R i_s=i_L+i_R Therefore, i R = i s − i L i_R=i_s-i_L Hence, i ˙ L =...

Start with a simple baseband tone Take m ( t ) = cos ⁡ ( ω m t ) m(t)=\cos(\omega_m t) and carrier cos ⁡ ( ω c t ) . \cos(\omega_c t). If we simply multiply them: s ( t ) = m ( t ) cos ⁡ ( ω c t ) s(t)=m(t)\cos(\omega_ct) then s ( t ) = cos ⁡ ( ω m t ) cos ⁡ ( ω c t ) . s(t)=\cos(\omega_mt)\cos(\omega_ct). Using cos ⁡ A cos ⁡ B = 1 2 [ cos ⁡ ( A + B ) + cos ⁡ ( A − B ) ] , \cos A\cos B =\frac12[\cos(A+B)+\cos(A-B)], we get s ( t ) = 1 2 cos ⁡ ( ( ω c + ω m ) t ) + 1 2 cos ⁡ ( ( ω c − ω m ) t ) \boxed{s(t)=\frac12\cos((\omega_c+\omega_m)t)+\frac12\cos((\omega_c-\omega_m)t)} So we get two frequencies : ω c + ω m \boxed{\omega_c+\omega_m} and ω c − ω m . \boxed{\omega_c-\omega_m}. These are the upper sideband (USB) and lower sideband (LSB) . We want only one Suppose we want the upper side...

  MATLAB Code clear; close all; clc; %% Parameters Fs = 1000;          % Sampling frequency T  = 2;             % Duration t  = 0:1/Fs:T-1/Fs; fm = 5;             % Message frequency fc = 100;            % Carrier frequency %% Baseband signal m = cos(2*pi*fm*t); %% Hilbert transform mh = imag(hilbert(m)); %% Analytic signal ma = m + 1j*mh; %% Envelope and instantaneous phase envelope = abs(ma); phase = unwrap(angle(ma)); %% I/Q modulation I = m; Q = mh; USB = I .* cos(2*pi*fc*t) ...     - Q .* sin(2*pi*fc*t); LSB = I .* cos(2*pi*fc*t) ...     + Q .* sin(2*pi*fc*t); %% Ordinary DSB-SC DSB = m .* cos(2*pi*fc*t); %% Plot everything figure('Color','w'); subplot(4,2,1) plot(t,m,'b') grid on title('Baseband m(t)') xlabel('Time (s)') subplot(4,2,2) plot(t,mh,'r') grid on title('Hilbert transform m̂(t)') xlabel('Time (s)') subplot(4,2,3)...

  Hilbert Transform & I/Q Simulator Hilbert Transform / Analytic Signal / I-Q Simulator Message frequency Carrier frequency Sampling frequency Duration Run simulation 1. Hilbert Transform m(t) = cos(2π fₘ t) m̂(t) = Hilbert{m(t)} 2. Analytic Signal mₐ(t) = m(t) + j m̂(t) Envelope = |mₐ(t)| 3. I/Q Modulation s(t) = I(t)cos(ωₙt) − Q(t)sin(ωₙt) I(t) = m(t) Q(t) = m̂(t) 4. SSB USB = m(t)cos(ωₙt) − m̂(t)sin(ωₙt) LSB = m(t)cos(ωₙt) + m̂(t)sin(ωₙt) 5. Frequency Domain Return to Premium Virtual DSP Lab →

The Hilbert transform of a signal x ( t ) is x ^ ( t ) = H { x ( t ) } = 1 π PV ∫ − ∞ ∞ x ( τ ) t − τ d τ \boxed{\hat{x}(t)=\mathcal H\{x(t)\} =\frac{1}{\pi}\operatorname{PV}\int_{-\infty}^{\infty} \frac{x(\tau)}{t-\tau}\,d\tau} where PV means the Cauchy principal value , because the kernel 1 t − τ is singular at τ = t . Convolving a time-domain signal f ( t ) f(t) with the Hilbert transform kernel 1 π t \frac{1}{\pi t} pro...