# Band Pass Signal Sampling MCQ [Free PDF] – Objective Question Answer for Band Pass Signal Sampling Quiz

1. The frequency shift can be achieved by multiplying the bandpass signal as given in the equation x(t) = $$u_c (t) cos⁡2π F_c t-u_s (t) sin⁡2π F_c t$$ by the quadrature carriers cos[2πFct] and sin[2πFct] and lowpass filtering the products to eliminate the signal components of 2Fc.

A. True
B. False

It is certainly advantageous to perform a frequency shift of the bandpass signal by sampling the equivalent low pass signal. Such a frequency shift can be achieved by multiplying the bandpass signal as given in the above equation by the quadrature carriers cos[2πFct] and sin[2πFct] and low pass filtering the products to eliminate the signal components at 2Fc.

Clearly, the multiplication and the subsequent filtering are first performed in the analog domain and then the outputs of the filters are sampled.

2. What is the final result obtained by substituting Fc=kB-B/2, T= 1/2B and say n = 2m i.e., for even and n=2m-1 for odd in equation x(nT)= $$u_c (nT)cos⁡2πF_c nT-u_s (nT)sin⁡ 2πF_c nT$$?

A. $$(-1)^m u_c (mT_1)-u_s$$

B. $$u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}$$

C. $$(-1)^m u_c (mT_1)- u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}$$

D. None

$$x(nT)=u_c (nT)cos⁡ 2πF_c nT-u_s (nT)sin⁡2πF_c nT$$ → equ1

=$$u_c (nT)cos⁡\frac{πn(2k-1)}{2}-u_s(nT)sin⁡\frac{πn(2k-1)}{2}$$ → equ2

On substituting the above values in equ1, we get say n=2m, $$x(2mT) ≡ xmT_{(1)} = u_c (mT_1)cos⁡πm(2k-1)=(-1)^m u_c (mT_1)$$

where $$T_1=2T=\frac{1}{B}$$. For n odd, say n=2m-1 in equ2 then we get the result as follows

$$u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}$$

Hence proved.

3. Which low pass signal component occurs at the rate of B samples per second with even-numbered samples of x(t)?

A. uc-lowpass signal component
B. us-lowpass signal component
C. uc & us-lowpass signal component
D. none of the mentioned

With the even-numbered samples of x(t), which occur at the rate of B samples per second, produce samples of the low pass signal component uc.

4. Which low pass signal component occurs at the rate of B samples per second with odd-numbered samples of x(t)?

A. uc – lowpass signal component
B. us – lowpass signal component
C. uc & us – lowpass signal component
D. none of the mentioned

With the odd-numbered samples of x(t), which occur at the rate of B samples per second, produce samples of the low pass signal component us.

5. What is the reconstruction formula for the bandpass signal x(t) with samples taken at the rate of 2B samples per second?

A. $$\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t-mT)$$

B. $$\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t-mT)$$

C. $$\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t+mT)$$

D. $$\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t+mT)$$

$$\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t-mT)$$, where T=1/2B

6. What is the new center frequency for the increased bandwidth signal?

A. Fc‘= Fc+B/2+B’/2
B. Fc‘= Fc+B/2-B’/2
C. Fc‘= Fc-B/2-B’/2
D. None of the mentioned

A new center frequency for the increased bandwidth signal is

Fc‘ = Fc+B/2-B’/2

7. According to the sampling theorem for low pass signals with T1=1/B, then what is the expression for uc(t) = ?

A. $$\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(π/T_1)(t-mT_1)}$$

B. $$\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1+T_1/2)}{(\frac{π}{T_1})(t-mT_1+\frac{T_1}{2})}$$

C. $$\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t+mT_1)}{(\frac{π}{T_1})(t+mT_1)}$$

D. $$\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t+mT_1+\frac{T_1}{2})}{(\frac{π}{T_1})(t+mT_1+\frac{T_1}{2})}$$

To reconstruct the equivalent low pass signals. Thus, according to the sampling theorem for low pass signals with T1=1/B.
$$u_c (t)=\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(π/T_1)(t-mT_1)}$$.

8. According to the sampling theorem for low pass signals with T1=1/B, then what is the expression for us(t) = ?

A. $$\sum_{m=-∞}^∞ u_c (mT_1) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(\frac{π}{T_1})(t-mT_1)}$$

B. $$\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1+\frac{T_1}{2})}{(π/T_1)(t-mT_1+\frac{T_1}{2})}$$

C. $$\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1-\frac{T_1}{2})}{(\frac{π}{T_1})(t-mT_1-\frac{T_1}{2})}$$

D. $$\sum_{m=-∞}^∞ u_c (mT_1) \frac{sin⁡(\frac{π}{T_1}) (t+mT_1)}{(\frac{π}{T_1})(t+mT_1)}$$

To reconstruct the equivalent low pass signals. Thus, according to the sampling theorem for low pass signals with T1=1/B .

$$u_s (t)=\sum_{m=-∞}^∞ u_s (mT_1-T_1/2) \frac{sin⁡(π/T_1) (t-mT_1+T_1/2)}{(π/T_1)(t-mT_1+T_1/2)}$$

9. What is the expression for low pass signal component uc(t) that can be expressed in terms of samples of the bandpass signal?

A. $$\sum_{n=-∞}^∞ (-1)^{n+r+1} x(2nT^{‘}-T^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘}+T^{‘})}{(π/(2T^{‘}))(t-2nT^{‘}+T^{‘})}$$

B. $$\sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘})}{(π/(2T^{‘}))(t-2nT^{‘})}$$

C. All of the mentioned

D. None of the mentioned

The low pass signal components uc(t) can be expressed in terms of samples of the
band pass signal as follows:

$$u_c (t) = \sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘})}{(π/(2T^{‘}))(t-2nT^{‘})}$$.

10. What is the expression for low pass signal component us(t) that can be expressed in terms of samples of the bandpass signal?

A. $$\sum_{n=-∞}^∞ (-1)^{n+r+1} x(2nT^{‘}-T^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘}+T^{‘})}{(π/(2T^{‘}))(t-2nT^{‘}+T^{‘})}$$

B. $$\sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘})}{(π/(2T^{‘}))(t-2nT^{‘})}$$

C. All of the mentioned

D. None of the mentioned

The low pass signal components us(t) can be expressed in terms of samples of the band pass signal as follows:

$$u_s (t) = \sum_{n=-∞}^∞ (-1)^{n+r+1} x(2nT^{‘}-T^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘}+T^{‘})}{(π/(2T^{‘}))(t-2nT^{‘}+T^{‘})}$$

11. What is the Fourier transform of x(t)?

A. X (F) = $$\frac{1}{2} [X_l (F-F_C.+X_l^* (F-F_C.]$$

B. X (F) = $$\frac{1}{2} [X_l (F-F_C.+X_l^* (F+F_C.]$$

C. X (F) = $$\frac{1}{2} [X_l (F+F_C.+X_l^* (F-F_C.]$$

D. X (F) = $$\frac{1}{2} [X_l (F-F_C.+X_l^* (-F-F_C.]$$

X (F) = $$\int_{-\infty}^∞ x(t)e^{-j2πFt} dt$$

=$$\int_{-\infty}^∞ \{Re[x_l (t) e^{j2πF_c t}]\}e^{-j2πFt} dt$$

Using the identity, Re(ε)=1/2(ε+ε^*)

X (F) = $$\int_{-\infty}^∞ [x_l (t) e^{j2πF_c t}+x_l^* (t)e^{-j2πF_c t}] e^{-j2πFt} dt$$

=$$\frac{1}{2}[X_l (F-F_C.+X_l^* (-F-F_C.]$$.

12. What is the basic relationship between the spectrum of the real bandpass signal x(t) and the spectrum of the equivalent low pass signal xl(t)?

A. X (F) = $$\frac{1}{2} [X_l (F-F_C.+X_l^* (F-F_C.]$$

B. X (F) = $$\frac{1}{2} [X_l (F-F_C.+X_l^* (F+F_C.]$$

C. X (F) = $$\frac{1}{2} [X_l (F+F_C.+X_l^* (F-F_C.]$$

D. X (F) = $$\frac{1}{2} [X_l (F-F_C.+X_l^* (-F-F_C.]$$

X(F) = $$\frac{1}{2} [X_l (F-F_C.+X_l^* (-F-F_C.]$$, where Xl(F) is the Fourier transform of xl(t). This is the basic relationship between the spectrum o f the real band pass signal x(t) and the spectrum of the equivalent low pass signal xl(t).

13. Which of the following is the right way of representing of the equation that contains only the positive frequencies in a given x(t) signal?

A. X+(F)=4V(F)X(F)
B. X+(F)=V(F)X(F)
C. X+(F)=2V(F)X(F)
D. X+(F)=8V(F)X(F)

In a real-valued signal x(t), has a frequency content concentrated in a narrow band of frequencies in the vicinity of a frequency Fc. Such a signal which has only positive frequencies can be expressed as X+(F)=2V(F)X(F)

Where X+(F) is a Fourier transform of x(t) and V(F) is the unit step function.

14. What is the equivalent time –domain expression of X+(F)=2V(F)X(F)?

A. F(+1)[2V(F)]*F(+1)[X(F)]
B. F(-1)[4V(F)]*F(-1)[X(F)]
C. F(-1)[V(F)]*F(-1)[X(F)]
D. F(-1)[2V(F)]*F(-1)[X(F)]

Given Expression, X+(F)=2V(F)X(F).It can be calculated as follows

$$x_+ (t)=\int_{-∞}^∞ X_+ (F)e^{j2πFt} dF$$

=$$F^{-1} [2V(F)]*F^{-1} [X(F)]$$

15. In time-domain expression, $$x_+ (t)=F^{-1} [2V(F)]*F^{-1} [X(F)]$$. The signal x+(t) is known as

A. Systematic signal
B. Analytic signal
C. Pre-envelope of x(t)
D. Both Analytic signal & Pre-envelope of x(t)

From the given expression, $$x_+ (t)=F^{-1} [2V(F)] * F^{-1}[X(F)]$$.

16. In equation $$x_+ (t)=F^{-1} [2V(F)]*F^{-1} [X(F)]$$, if $$F^{-1} [2V(F)]=δ(t)+j/πt$$ and $$F^{-1} [X(F)]$$ = x(t). Then the value of ẋ(t) is?

A. $$\frac{1}{π} \int_{-\infty}^\infty \frac{x(t)}{t+τ} dτ$$

B. $$\frac{1}{π} \int_{-\infty}^\infty \frac{x(t)}{t-τ} dτ$$

C. $$\frac{1}{π} \int_{-\infty}^\infty \frac{2x(t)}{t-τ} dτ$$

D. $$\frac{1}{π} \int_{-\infty}^\infty \frac{4x(t)}{t-τ} dτ$$

$$x_+ (t)=[δ(t)+j/πt]*x(t)$$

$$x_+ (t)=x(t)+[j/πt]*x(t)$$

$$ẋ(t)=[j/πt]*x(t)$$

=$$\frac{1}{π} \int_{-\infty}^\infty \frac{x(t)}{t-τ} dτ$$ Hence proved.

17. If the signal ẋ(t) can be viewed as the output of the filter with impulse response h(t) = 1/πt, -∞ < t < ∞ when excited by the input signal x(t) then such a filter is called as __________

A. Analytic transformer
B. Hilbert transformer
C. Both Analytic & Hilbert transformer
D. None of the mentioned

The signal ẋ(t) can be viewed as the output of the filter with impulse response h(t) = 1/πt,

-∞ < t < ∞ when excited by the input signal x(t) then such a filter is called a Hilbert transformer.

18. What is the frequency response of a Hilbert transform H(F)=?

A. $$\begin{cases}&-j (F>0) \\ & 0 (F=0)\\ & j (F<0)\end{cases}$$

B. $$\left\{\begin{matrix}-j & (F<0)\\0 & (F=0) \\j & (F>0)\end{matrix}\right.$$

C. $$\left\{\begin{matrix}-j & (F>0)\\0 &(F=0) \\j & (F<0)\end{matrix}\right.$$

D. $$\left\{\begin{matrix}j&(F>0)\\0 & (F=0)\\j & (F<0)\end{matrix}\right.$$

H(F) =$$\int_{-∞}^∞ h(t)e^{-j2πFt} dt$$

=$$\frac{1}{π} \int_{-∞}^∞ 1/t e^{-2πFt} dt$$

=$$\left\{\begin{matrix}-j& (F>0)\\0&(F=0) \\ j& (F<0)\end{matrix}\right.$$

We Observe that │H (F)│=1 and the phase response ⊙(F) = -1/2π for F > 0 and ⊙(F) = 1/2π for F < 0.

19. What is the equivalent lowpass representation obtained by performing a frequency translation of X+(F) to Xl(F)= ?

A. X+(F+FC.
B. X+(F-FC.
C. X+(F*FC.
D. X+(Fc-F)

The analytic signal x+(t) is a bandpass signal. We obtain an equivalent lowpass representation by performing a frequency translation of X+(F).

20. What is the equivalent time domain relation of xl(t) i.e., lowpass signal?

A. $$x_l (t)=[x(t)+j ẋ(t)]e^{-j2πF_c t}$$

B. x(t)+j ẋ(t) = $$x_l (t) e^{j2πF_c t}$$

C. $$x_l (t)=[x(t)+j ẋ(t)]e^{-j2πF_c t}$$ & x(t)+j ẋ(t) = $$x_l (t) e^{j2πF_c t}$$

D. None of the mentioned

$$x_l (t)=x_+ (t) e^{-j2πF_c t}$$

=$$[x(t)+j ẋ(t)] e^{-j2πF_c t}$$

Or equivalently, x(t)+j ẋ(t) =$$x_l (t) e^{j2πF_c t}$$.

21. If we substitute the equation $$x_l (t)= u_c (t)+j u_s (t)$$ in equation x (t) + j ẋ (t) = xl(t) ej2πFct and equate real and imaginary parts on side, then what are the relations that we obtain?

A. x(t)=$$u_c (t) \,cos⁡2π \,F_c \,t+u_s (t) \,sin⁡2π \,F_c \,t$$; ẋ(t)=$$u_s (t) \,cos⁡2π \,F_c \,t-u_c \,(t) \,sin⁡2π \,F_c \,t$$

B. x(t)=$$u_c (t) \,cos⁡2π \,F_c \,t-u_s (t) \,sin⁡2π \,F_c \,t$$; ẋ(t)=$$u_s (t) \,cos⁡2π \,F_c t+u_c (t) \,sin⁡2π \,F_c \,t$$

C. x(t)=$$u_c (t) \,cos⁡2π \,F_c t+u_s (t) \,sin⁡2π \,F_c \,t$$; ẋ(t)=$$u_s (t) \,cos⁡2π \,F_c t+u_c (t) \,sin⁡2π \,F_c \,t$$

D. x(t)=$$u_c (t) \,cos⁡2π \,F_c \,t-u_s (t) \,sin⁡2π \,F_c \,t$$; ẋ(t)=$$u_s (t) \,cos⁡2π \,F_c \,t-u_c (t) \,sin⁡2π \,F_c \,t$$

If we substitute the given equation with another, then we get the required result.

22. In the relation, x(t) = $$u_c (t) cos⁡2π \,F_c \,t-u_s (t) sin⁡2π \,F_c \,t$$ the low frequency components uc and us are called _____ of the bandpass signal x(t).