Solution (Answer Option 2) s(s^2+s+1)(s+4) + k = s^4+5s^3+5s^2+4s+K Applying Routh-Hurwitz Criterion,  S^4     1       5      K S^3     5       4      0 S^2     21/5  K     0 S        (84/5 - 5K)/(21/5) 0  0 S^0     K (84/5 - 5K)/(21/5) or (84 - 25K)/21 should be greater than 0  Or,  84 − 25 K > 0 K < 84 25 K < 25 84 ​ and K should be greater than 0 Or,  0 <  K < 25 84 ​

  Step 1: Calculate noise power Given: B = 1   M H z = 10 6   H z B=1\ MHz=10^6\ Hz Noise spectral density = − 174   d B m / H z =-174\ dBm/Hz Signal power S = − 80   d B m S=-80\ dBm Total noise power: N = − 174 + 10 log ⁡ 10 ( 10 6 ) N=-174+10\log_{10}(10^6) N = − 174 + 60 = − 114   d B m N=-174+60=-114\ dBm Step 2: Calculate SNR S N R = S − N SNR=S-N = − 80 − ( − 114 ) =-80-(-114) S N R = 34   d B \boxed{SNR=34\ dB} Convert into linear scale: S N R = 10 34 / 10 = 2511.89 SNR=10^{34/10}=2511.89 Step 3: Calculate Shannon capacity Shannon-Hartley theorem: C = B log ⁡ 2 ( 1 + S N R ) C=B\log_2(1+SNR) Substitute: C = 10 6 log ⁡ 2 ( 2512.89 ) C=10^6\log_2(2512.89) C = 11.295 × 10 6 C=11.295\times10^6 Therefore, C = 11.295   M b p s \boxed{C=11.295\ Mbps} Step 4: Compare with the options Since B = 1   M H z B=1\ MHz : Option Condition Result A C = B C=B False B C = 2 B C=2B False C C > 3 B C>3B True D C < B C...

  Q. Consider an LED based on a direct bandgap semiconductor material with energy bandgap 1.3 eV. Given: Plank’s constant, ℎ=6.63×10^−34 J s and speed of light in free space is 3×108 m s^−1. In which of the following wavelength ranges the LED will NOT emit? Given Data Bandgap energy: Eg = 1.3 eV Planck's constant: h = 6.63 × 10⁻³⁴ J s Speed of light: c = 3 × 10⁸ m/s Step 1: Convert Bandgap Energy into Joules We know that: 1 eV = 1.602 × 10⁻¹⁹ J Therefore, Eg = 1.3 × 1.602 × 10⁻¹⁹ Eg = 2.0826 × 10⁻¹⁹ J Step 2: Apply Planck's Formula Photon energy is given by: E = hc / λ λ = hc / E Substituting the values: λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (2.0826 × 10⁻¹⁹) λ = 9.55 × 10⁻⁷ m Step 3: Convert Metres in...

  Step-by-Step Solution Find the correct value of the current I x . Step 1: Apply the Virtual Ground Concept For an ideal op-amp with negative feedback: The non-inverting terminal (+) is grounded. Therefore, V - = V + = 0 V. No current enters the op-amp input terminals. Step 2: Apply KCL at the Inverting Terminal Let the voltage at the right-hand node be V x . Applying Kirchhoff's Current Law: (2 - 0) / 1k = (0 - V x ) / 1k Therefore, V x = -2 V Step 3: Apply KCL at the Right-Hand Node First, calculate the current through the feedback 1 kΩ resistor: I 1 = [0 - (-2)] / 1k I 1 = 2 mA Next, calculate the curre...

  Q. A complex load (in Ω) is represented as Γ L = 0.5∠30° on the Smith chart. A co-axial cable with a characteristic impedance of 50 Ω is connected to the load. The new input impedance of the load now moves to a diametrically opposite point on the same Γ circle on the Smith chart. Which option is the nearest input impedance of the cable-connected load? Options (A) 20.7 − j5.1 Ω (B) 17.7 − j11.8 Ω ✓ (C) 97.5 − j65.0 Ω (D) 97.5 + j65.0 Ω Step-by-Step Solution Step 1: Given Data Load reflection coefficient: Γ L = 0.5∠30° Characteristic impedance: Z 0 = 50 Ω Step 2: Find the New Reflection Coefficient A diametrically opposite point on the same VSWR circl...

  Q. A small signal source, V i (t) = A cos(10 5 t) + B sin(10 7 t) , is applied to a BJT circuit as shown in the figure. Assume zero source resistance, V BE = 0.7 V, β dc = 99, Early voltage = 100 V and thermal voltage = 25 mV. The effect of internal parasitic capacitances of the BJT may be neglected. Which expression is the best approximation of the output voltage V o (t)? Step-by-step Solution Step 1: Calculate the DC Operating Point Given circuit parameters: V CC = 12 V R 1 = 100 kΩ, R 2 = 25 kΩ R C = 5 kΩ R E = 500 + 1000 = 1500 Ω β = 99, V BE = 0.7 V Thevenin equivalent voltage at the base: V TH = 12 × 25 / (100 + 25) = 2.4 V Thevenin resistance: ...