Q1
A 500 MHz signal with 50 mVpeak-peak amplitude must be converted by an N-bit ADC.
The sample clock shows a jitter of 1 psrms, sampling on a 2 pF sample capacitor.
The second order harmonic distortion of the signal is at -60 dB.
An ENOB=7 bit is required.
A
Calculate N, round up to integer.
B
The signal amplitude is now increased.
All values (jitter, sample cap, distortion) remain the same and use N=8.
At what amplitude is ENOB=7.4 bit reached?
Q2
The output signal of an FM intermediate frequency circuit has a bandwidth of 100 kHz at a carrier frequency of 10.7 MHz.
An analog-to-digital converter samples this signal at 5.35 Ms/s.
What resolution is needed to obtain a SNQR due to quantization of 13.8 bit in 100 kHz bandwidth.
Q3
An ideal 10 bit ADC is sampling at 50 Ms/s a full-scale 12 MHz input signal with 2nd harmonic.
The second harmonic is expected worst case to be around 54 dB level.
A
What is the expected SNDR?
B
The measurement shows an SNDR of 56 dB.
At what level is the 2nd harmonic?
C
In order to measure the 2nd harmonic exactly, a filter around
What bandwidth is required?
D
There are only low-pass filters (from DC to BW) available.
How can you measure the 12 MHz signal and its harmonic?
Q4
An 10-bit ADC samples at 200 Ms/s an input signal of 500 mVpp in a band between 80-90MHz.
The input uses a 3 pF capacitor and the jitter on the sampling signal is 1 psrms.
A
What is the effective resolution (ENOB) in the 10 MHz band between 80 MHz and 90 MHz?
B
The succeeding digital electronics uses only every second sample and discards the intermediate samples, thereby halving the sample rate.
What will happen to the signal and what is the ENOB in the resulting 10 MHz bandwidth?
C
Is it a good idea to reduce the sample rate of the ADC to 100 Ms/s?
D
Is there a better solution resulting in a higher ENOB?