(b) The block diagram in Figure Q3B describes a measurement system based on a lock-in amplifier configuration. An invert
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(b) The block diagram in Figure Q3B describes a measurement system based on a lock-in amplifier configuration. An invert
(b) The block diagram in Figure Q3B describes a measurement system based on a lock-in amplifier configuration. An inverting amplifier is used as the input stage, where R1 = 2 k1, R2 = 20 k and Av = 1,000. The amplification stage has a gain bandwidth product (GBP) of 10^6 and 0.1 mV/Hz of input noise. R w Bandpass Lowpass Mixer Vbp TO ADC Vout Vpsd Vdc Vref a filter filter w AR VN Figure Q3B i) Derive the exact algebraic equation that describes the output of the first amplifier stage Vout. ii) Calculate the exact value of the inverting amplifier closed loop gain for an operating frequency of 100 Hz. EE312 Page 9 of 12 iii) Calculate the useful bandwidth of the inverting amplifier. The input signal Vin is a sine wave signal which has amplitude 10 mv, a carrier frequency of 90 kHz and phase 0. The reference signal Vret is a sine wave signal which has amplitude 2V, a carrier frequency of 90 kHz and phase 0. Before the signal can be processed by the PSD (or Mixer), a bandpass filter is used to reduce the noise at the output Vout of the amplification stage. The bandpass filter transfer function has a gain BP = 1 and a bandwidth of 5000 Hz centred around 90 kHz. iv) Calculate the signal-to-noise ratio (SNR) of the signal Vbp at the output of the bandpass filter. (Hint: you should consider that SNR = 10 logo ARMS-5 ARMS-N 2010g10 12 Vpoisel v) Derive the EXACT equation that describes the signal Vpsd (at the output of the mixer) in the case of a noiseless input. Remember that: A sin(ot+)* A, sin(0,1+2) = 1 2 1 = 54,4, cos(0) - 6,5+0,-0.)- 24/4, cos(0) + ®,]+0,+0.) = -0 After the mixer, a low-pass filter is present with gain LP = 1, as represented in Figure Q3B. V 双击看大图 bandwidth of the low pass filter in order to obtain a SNR = 10 dB at the output of the lowpass filter.
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