Final Answer:
The expression for V1(s)/Vt(s) in the given circuit is V1(s)/Vt(s) = R2 / (R1 + R2) / (1 + s * R2 * C1). This transfer function represents a low-pass STC (Series-Trap-Capacitor) filter. For R1 = 10k ohm, R2 = 40k ohm, and C1 = 5 pF, the 3-dB frequency is approximately 795.77 kHz.
Step-by-step explanation:
In the circuit given with the signal source connected to the amplifier, the transfer function V1(s)/Vt(s) (where V1(s) is the output voltage and Vt(s) is the input voltage) can be derived using the voltage divider rule. The voltage across R2 (output) divided by the total voltage is expressed as V1(s)/Vt(s) = R2 / (R1 + R2) / (1 + s * R2 * C1), considering R1 as the source resistance, R2 as the input resistance of the amplifier, and C1 as the input capacitance of the amplifier.
This transfer function represents a low-pass filter, as it attenuates higher frequencies and allows lower frequencies to pass. The expression indicates a standard form of a low-pass filter where the cutoff frequency, also known as the 3-dB frequency, can be calculated by setting the magnitude of the transfer function to 1/√2 and solving for the frequency. For the given values of R1 = 10k ohm, R2 = 40k ohm, and C1 = 5 pF, the 3-dB frequency is approximately 795.77 kHz, representing the frequency at which the output power drops to half of its maximum value.
In conclusion, the derived transfer function V1(s)/Vt(s) represents a low-pass STC filter, and by calculating the 3-dB frequency for specific values of resistances and capacitance, it's evident that the circuit exhibits a cutoff frequency around 795.77 kHz, confirming its low-pass filtering characteristics.