Group Problem Solving 915MHz Receiver Front End Experiment -30 35 -40 45 3-50 RF
ID: 2250071 • Letter: G
Question
Group Problem Solving 915MHz Receiver Front End Experiment -30 35 -40 45 3-50 RF Pre-amp Miteq IF AMP ZFL10OOLN ZEM4300 70MHz Sq. Wave 915MHz BPF PTI 70 MHz BPF S BP-70 Modulated Signal to SA and Scope -60 865 880 895 910 925 940 955 Cushcraft LO) VCO Frequency (MHz) fo = 845 MHz Z05-1025 If the LO signal in the above is changed to 845 MHz, and it is desired to reject the image frequency and interfering signals at 890 (and below) and 930 MHz (and above), specify the (maximum) bandwidths required, of A. The RF filter ( design this to reject the image frequency) B. The IF filter. (design this to reject the close in interferers) In doing so you can assume for our purposes that the filters have infinite rejection outside the passband to keep things simple 16Explanation / Answer
Bandwidth is generally the difference between the higher frequency and the lower frequency.
A. In case of RF filter,the image frequency must be rejected.
The bandwidth for RF filter is given as Bw=915-fLo
=915-845
=70db
B. In case of IF filter,the close in interferers must be rejected.
The Bw of IF filter=930-845 (as 845 is the lower frequency, 70Hz cannot be considered as higher frequency. Hence in this case,930 Hz is taken)
=85db
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