Ask question

Ask question

All categories

3 years ago

9

Consider a continuous-time LTI system whose frequency response is I x sin(4w) H(jw) = -x h(t)e-jwtdt = w If the input to this sy

stem is a periodic signal x(t) = { ~ l, O:s:t<4 4:St<8 with period T = 8, determine the corresponding system output y(t).

1 answer:

Helga [31]3 years ago

8 0

Answer:

I believe this is your question.

See the hand worked solution attached.

You might be interested in

Technician A says if the input signal turn-on time is too fast for the input circuit, your program may operate as though the inp

ASHA 777 [7]

Answer:

b. Technician B only

Explanation:

A watchdog timer is a circuit that automatically monitors the MCU (Microcontroller Unit) for any anomaly, detects it and helps the MCU to recover from the malfunction it has detected.

If the input signal turn-on time is too fast for the input circuit, that is a malfunction and this activates the watchdog timer circuit to correct this malfunction immediately. So Technician B only is correct as the watchdog timer is activated immediately once there is a malfunction.

7 0

3 years ago

5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of

marta [7]

Answer:

(a) $3.185*10^{21} cells$

(b) $6.37*10^{21} atoms$

Explanation:

(a)

Volume, V of unit cell

$V=(2.866*10^{-8})^{3}=2.354*10^{-23}$

Number of unit cells, N

$N=\frac {W_{mat}}{V\rho_{mat}}$ Where $W_{mat}$ is weight of material and $\rho_{mat}$ is density of material

$N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells$

(b)

Number of atoms in paper clip

This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

Number of atoms of iron= $3.185*10^{21} cells*2 atoms/cell=6.37*10^{21} atoms$

8 0

3 years ago

In your Reader/Writer Notebook, write a short first-person narrative from the perspective of the bank clerk, describing her phys

guajiro [1.7K]

BAHHHHAHHH HOFHOFYOCIGC

5 0

3 years ago

A sample of sand weighs 490 g in stock and 475 in Oven Dry (OD) condition, respectively. If absorption capability of the sand is

Ivahew [28]

The weight of the specimen in SSD condition is 373.3 cc

<u>Explanation</u>:

a) Apparent specific gravity = $\frac{A}{A-C}$

Where,

A = mass of oven dried test sample in air = 1034 g

B = saturated surface test sample in air = 1048.9 g

C = apparent mass of saturated test sample in water = 975.6 g

apparent specific gravity = $\frac{A}{A-C}$

= $\frac{1034}{1034-675 \cdot 6}$

Apparent specific gravity = 2.88

b) Bulk specific gravity $G_{B}^{O D}=\frac{A}{B-C}$

$G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}$

= 2.76

c) Bulk specific gravity (SSD):

$G_{B}^{S S D}=\frac{B}{B-C}$

$=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}$

$G_{B}^{S S D}$ = 2.80

d) Absorption% :

$=\frac{B-A}{A} \times 100 \%$

$=\frac{1048 \cdot 9-1034}{1034} \times 100$

Absorption = 1.44 %

e) Bulk Volume :

$v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}$

$=\frac{1048 \cdot 9-675 \cdot 6}{1}$

= $373.3 cc$

5 0

2 years ago

A discrete MOSFET common-source amplifier has RG = 2 MΩ, gm = 5 mA/V, ro = 100 kΩ, RD = 20kΩ, Cgs = 3pF, and Cgd = 0.5pF. The am

Papessa [141]

Answer:

a) -36.36 V/V

b) 15.17 kHz

c) 1.6 GHz

Explanation:

See attached picture.

7 0

3 years ago