Unified Filter & Q-Factor Analyzer Exploring Gain, Selectivity, ripples, and Side-Lobes Design Relationship Q = f c / BW BW = -- Hz Transfer Function Characteristic |H(f)| ≈ 1 / √(1 + ε 2 C n 2 ) Control Panel Filter Type: Butterworth (Flat / No Lobes) Chebyshev (Equiripple / Side-Lobes) Center Frequency: 1000 Hz Quality Factor (Q): 5.0 Ripple/Lobe Intensity: 0.5 Recommended Design Multiple Feedba...

100. Which of the following statements are correct for active filters? A. Multiple-feedback are preferred for Q ≤ 10. B. For Q > 10, state-variable filter is used. C. The upper and lower cut-off frequencies have arithmetic symmetry about centre frequency f c for smaller Q values. D. As Q is increased, the filter becomes less selective and both upper and lower cut-off frequencies approximate geometric symmetry. Choose the correct answer from the options given below: A and B only [Option ID = 1229] A and C only [Option ID = 1230] B and C only [Option ID = 1231] C and D only [Option ID = 1232] Correct Answer: 1 Solution: Statement A: Correct. Multiple-feedback (MFB) circuits are simple but get unstable if the "Q" (sharpness) is too high. They are be...

  99. The characteristic equation for the output voltage of an All-pass filter is given by: V o = V in [ 1 - 2 / (jωRC + 1) ] [Option ID = 1225] V o = V in [ 1 - 2 / (jωRC) ] [Option ID = 1226] V o = V in [ -1 + 2 / (jωRC + 1) ] [Option ID = 1227] V o = V in [ 2 / (jωRC + 1) ] [Option ID = 1228] Correct Answer: 1 Solution Based on the question, the correct characteristic equation is Option 1 : V o = V in [ 1 - 2 / (jωRC + 1) ] Comparison: How others look To identify filters quickly, look at the numerator (the top part of the fraction): 1. Low-Pass Filter (Blocks Highs) Notice: It has only a "1" on top. It is the simplest one. V o = V in [ 1 / (jωRC + 1) ] 2. High-Pass Filter (Blocks Lows) Notice: It has the frequency symbol (jωRC) on top. V o = V in [ j...

  98. A filter is needed to remove induced 60 Hz hum from a transducer signal. The rejection bandwidth is 2 Hz. The correct combination of R damping and C values for this filter are: 1. 30 Ω, 32.5 milli Farads [Option ID = 3221]  2. 68 Ω, 30.5 milli Farads [Option ID = 3222]  3. 82 Ω, 3.25 milli Farads [Option ID = 3223]  4. 89 Ω, 2.65 milli Farads [Option ID = 3224]  Correct Answer : 4 Solution To remove the 60 Hz frequency, we use a notch filter with a center frequency (f 0 ) of 60 Hz and a rejection bandwidth (BW) of 2 Hz. Step 1: Calculate the Quality Factor (Q) The Q factor determines the sharpness of the notch: Q = f 0 / BW = 60 / 2 = 30 Step 2: Calculate the Damping Resistance (R damping ) Using the damping relationship for this specific filter topology: R damping = 3Q - 1 = 3(30) - 1 = 89 Ω Step 3: Calculate the...

  97. Which of the following statements is NOT correct about Butterworth filter?  1. Butterworth is the most popular alignment type. [Option ID = 3217]  2. Butterworth is characterised by its moderate amplitude and phase response. [Option ID = 3218]  3. It exhibits the slowest roll‐off of any monoatomic (single‐slope) filter. [Option ID = 3219]  4. This is the only filter whose 3dB down frequency equals its critical frequency (f =f ). [Option ID = 3220]  Correct Answer : 3 Solution Butterworth filter has a slower roll-off compared to the Chebyshev or Elliptic filters, it does not have the slowest roll-off of all monotonic filters. The Bessel filter actually has a much slower roll-off than the Butterworth filter for the same order (n). The Butterworth is "middle-of-the-road" in terms of roll-off speed. Previous yr Question papers with Full Explanations → Electronics and Communiaction Study Materials → ...

  96. Which of the following statements is NOT correct for a Chebyshev Filter?  1. Chebyshev filter tends to exhibit ringing effect with transient signals. [Option ID = 3213]  2. Time delay and phase characteristics of Chebyshev filter are comparitively better than that of Butterworth filter. [Option ID = 3214]  3. Chebyshev filters have a sharper slope than Butterworth filters. [Option ID = 3215]  4. Chebyshev filters are capable of achieving more attenuation in stopband. [Option ID = 3216]  Correct Answer : B Solution Chebyshev filters achieve a sharper cutoff by allowing ripple (usually in the passband for Type I). This sharper response comes at the cost of poorer phase linearity and more variation in group delay than a Butterworth filter. Therefore, Butterworth filters generally have better phase and time-delay characteristics. Previous yr Question papers with Full Explanations → Electronics and Communi...

Simulation of BER vs. SNR for OFDM Modulation with BPSK Read the theory # OFDM Symbols: Subcarriers (N): Cyclic Prefix (Ncp): SNR (dB): Reset BER Results Log SNR (dB) BER Update BER vs SNR Plot Transmitted Bits (Partial) Received Bits (Partial) Compare the BER with Original BPSK → Mathematical Logic of the Simulation 1. Bit-to-Symbol Mapping (BPSK) Each bit \(b_i \in \{0, 1\}\) is mapped to a bipolar complex value: If bit = 1, \(X_k = 1 + 0j\) If bit...

Theoretical BER vs SNR for OFDM Theoretical BER vs SNR for OFDM Fundamental Rule: OFDM is a multiplexing scheme, not a modulation scheme. In an ideal AWGN channel, the BER of OFDM is identical to the BER of the underlying modulation used on its sub-carriers. If you use QPSK on your OFDM sub-carriers, your theoretical BER curve is simply the QPSK curve. 1. Standard Formulas (AWGN Channel) Assuming \(N\) sub-carriers and let \(\gamma_b = \frac{E_b}{N_0}\) (SNR per bit): BPSK / QPSK OFDM: \[P_b = Q\left(\sqrt{2 \gamma_b}\right) = \frac{1}{2}\text{erfc}\left(\sqrt{\gamma_b}\right)\] 16-QAM OFDM (Gray Coded): \[P_b \approx \frac{3}{4} Q\left(\sqrt{\frac{4}{5} \gamma_b}\right)\] 64-QAM OFDM: \[P_b \approx \frac{7}{12} Q\left(\sqrt{\frac{2}{7} ...