A common-emitter amplifier circuit is shown in Figure 3(a). Neglect r_x and r_o and assume the current source to be ideal. (i)…

(a) Common-Emitter Amplifier: Low-Frequency Response

Small-signal model. r_x and r_o are neglected, and the ideal current source is an open circuit to signals. So the emitter is returned to ground only through C_E. The collector sees R_C to ground (via V_CC), and R_L through C_C.

(i) Midband gain. In midband, C_E is a short (emitter grounded) and C_C is a short.

• The base–emitter voltage is v_π = V_sig·r_π/(R_sig + r_π).
• The output is V₀ = −g_m v_π(R_C ∥ R_L).

A_M = V₀/V_sig = −g_m(R_C ∥ R_L)·r_π/(R_sig + r_π) = −β(R_C ∥ R_L)/(R_sig + r_π)

(ii) Break frequencies. C_E sees the resistance looking into the emitter, R_E′ = (R_sig + r_π)/(β + 1) = r_e + R_sig/(β + 1). Hence:

ω_PE = 1/[C_E(R_sig + r_π)/(β + 1)]

C_C sees R_C + R_L in series, so:

ω_C = 1/[C_C(R_C + R_L)]

Because the current source has infinite resistance, the zero produced by C_E lies at s = 0, and so does the zero due to C_C.

(iii) Gain function. Exact analysis gives

A(s) = A_M·[s/(s + ω_PE)]·[s/(s + ω_C)]

Each capacitor contributes one pole and one zero at the origin.

(iv) Numerical midband gain. I = 1 mA, so g_m = I/V_T = 1/25 mV = 40 mA/V and r_π = β/g_m = 2.5 kΩ.

A_M = −100 × (10k ∥ 10k)/(10k + 2.5k) = −100 × 5k/12.5k = −40 V/V (32 dB)

(v) Capacitor selection.

• R_E′ = 12.5k/101 = 123.8 Ω
• R_C + R_L = 20 kΩ

Capacitance needed ∝ 1/(R·f). The higher break frequency (100 Hz, which sets f_L) should therefore go to the capacitor that sees the smaller resistance, C_E.

C_E = 1/(2π × 100 × 123.8) = 12.9 μF (f_PE = 100 Hz)

C_C = 1/(2π × 10 × 20 × 10³) = 0.80 μF (f_C = 10 Hz)

Total = 13.7 μF. The reverse choice would need 128.6 + 0.08 = 128.7 μF. The exact f_L comes out at 101 Hz, close to the 100 Hz target.

(vi) Bode plot.

• Above 100 Hz the gain is flat at 32 dB.
• Between 10 Hz and 100 Hz it falls at 20 dB/decade, reaching 12 dB (×4) at 10 Hz.
• Below 10 Hz it falls at 40 dB/decade.

Unity gain: 12 dB − 40 log(10/f) = 0 ⇒ f ≈ 5 Hz (exact curve: 5.3 Hz).

(b)(i) Routh-Hurwitz Test

Routh array for P(s) = s⁴ + 4s³ + 8s² + 12s + 15

• s⁴ row: 1, 8, 15
• s³ row: 4, 12
• s² row: (32 − 12)/4 = 5, 15
• s¹ row: (60 − 60)/5 = 0

Row of zeros. The s¹ row is entirely zero, which signals roots placed symmetrically about the origin. Form the auxiliary polynomial from the s² row:

A(s) = 5s² + 15, dA/ds = 10s

Replace the s¹ row with 10 and complete the array:

• s¹ row: 10
• s⁰ row: 15

Counting roots.

• The first column (1, 4, 5, 10, 15) has no sign change, so there are no roots with positive real part.
• A(s) = 0 gives s² = −3, so s = ±j√3. That is two roots with zero real part.
• The remaining two roots have negative real parts. Indeed P(s) = (s² + 3)(s² + 4s + 5), which gives s = −2 ± j.

Stability. The system has a non-repeated pair of poles on the jω-axis, so it is marginally stable. It will sustain an undamped oscillation at ω = √3 rad/s, and it is not asymptotically (BIBO) stable.

(b)(ii) Impulse Response of the Network

Circuit. v_in drives a 1 Ω series resistor into node v₀. From v₀ to ground there are two paths: the 1 F capacitor, and the 1 H inductor in series with 3 Ω.

Transfer function. KCL at v₀ in the s-domain, with zero initial conditions:

(V_in − V₀)/1 = sV₀ + V₀/(s + 3)

V₀[1 + s + 1/(s + 3)] = V_in

H(s) = V₀/V_in = (s + 3)/[(s + 1)(s + 3) + 1] = (s + 3)/(s² + 4s + 4) = (s + 3)/(s + 2)²

Partial fractions. Since s + 3 = (s + 2) + 1:

H(s) = 1/(s + 2) + 1/(s + 2)²

Impulse response. Using L⁻¹{1/(s + a)²} = te^(−at):

h(t) = (1 + t)e^(−2t) u(t) V (per unit-area voltage impulse)

The network is critically damped (repeated pole at s = −2, time constant 0.5 s). h(0⁺) = 1, because the impulse charges the 1 F capacitor through 1 Ω to 1 V, and h(t) decays to zero without oscillating.

(c) Slip and Speed at Maximum Torque

Given: 4 poles, 50 Hz, R₂ = 0.02 Ω, X₂ = 0.5 Ω. Stator impedance is neglected.

Synchronous speed: N_s = 120f/P = 120 × 50/4 = 1500 rpm

Slip for maximum torque. The motor is treated as a forward-rotating-field machine, where T ∝ (sE₂²R₂)/(R₂² + (sX₂)²). Setting dT/ds = 0 gives R₂ = s_m X₂, since R₁ = X₁ = 0:

s_m = R₂/X₂ = 0.02/0.5 = 0.04 (4 %)

Speed at maximum torque:

N = N_s(1 − s_m) = 1500 × (1 − 0.04) = 1440 rpm

The maximum torque itself does not depend on R₂; only the slip at which it occurs does. Adding rotor resistance would shift the peak towards standstill.

In a single-phase machine the backward field also acts. If the backward branch's reactance is kept in series (double-revolving-field model), the effective reactance roughly doubles and the peak moves to about s ≈ 0.02. The conventional answer asked for here is s_m = 0.04 and 1440 rpm.