设\(z = {e^{x - 2y}}\)，而\(x = \sin t,\;y = {t^3},\)则\( { { dz} \over {dt}} = \)( ) A: \({e^{\sin t - 2{t^3}}}\) B: \({e^{\sin t - 2{t^3}}}\left( {\cos t - 6{t^2}} \right)\) C: \({e^{\sin t - 2{t^3}}}\ {\sin t } \) D: \({e^{\sin t - 2{t^3}}}\,{t^3}\)

将正弦电压u = 10 sin( 314 t +30 ) V 施加于感抗XL = 5 的电感元件上,<br/>则通过该元件的电流 i = ( ) 。 A: 50 sin( 314 t +90 ) B: 2 sin( 314 t +60 ) C: 2 sin( 314 t -60 ) D: 2 sin( 314 t -30 )

设\(z = {e^{x - 2y}}\)，而\(x = \sin t\)，\(y = {t^3}\)，则全导数\( { { dz} \over {dt}} = \) A: \({e^{\sin t - {t^3}}}(\cos t - 6{t^2})\) B: \({e^{\sin t - 2{t^3}}}(\sin t - 6{t^2})\) C: \({e^{\cos t - 2{t^3}}}(\cos t - 6{t^2})\) D: \({e^{\sin t - 2{t^3}}}(\cos t - 6{t^2})\)

一阶常微分方程[img=152x26]1802e4d6075ee4f.png[/img]的通解为 A: sin(2*t)/5-cos(2*t)/10+C*exp(-4*t) B: sin(2*t)/7+cos(2*t)/5-C*exp(-3*t) C: sin(2*t)/7-C*cos(2*t)/10+C*exp(-2*t) D: sin(2*t)/7-cos(2*t)/7+C*exp(-5*t)

求微分方程[img=269x55]17da6536a9fba07.png[/img]的通解； （ ） A: (C15*sin(2*t))/exp(3*t) + (C16*sin(2*t))/exp(3*t) B: (C15*cos(2*t))/exp(3*t) - (C16*sin(2*t))/exp(3*t) C: (C15*cos(2*t))/exp(3*t) + (C16*cos(2*t))/exp(3*t) D: (C15*cos(2*t))/exp(3*t) + (C16*sin(2*t))/exp(3*t)

‏已知函数[img=102x27]18030256dad01f2.png[/img]，求其三阶导数，下面命令正确的是（）‍ A: syms t; G=simplify(diff(t^2*sin(t),t,3)) B: syms t; G=simplify(int(t^2*sin(t),t,3)) C: syms t; G=simplify(diff(t^2*sin(t),t)) D: syms t; G=simplify(int(t^2*sin(t),t))

用R程序编程求下列的值，正确的命令是 （ ）[img=76x44]180398c776266cb.jpg[/img] A: (sin(50*180/pi))^2/(1+ sin(10*180/pi)) B: (sin(50*π/180))^2/(1+ sin(10*π/180)) C: sin(50)^2/(1+ sin(10)) D: (sin(50*pi/180))^2/(1+ sin(10*pi/180))

在图2所示系统中，输入xc(t)=cos(2π5t)，若采样间隔T=1/16s，H([img=24x22]1802e7d54318646.png[/img])为理想全通系统，输出yc(t)=( )[img=818x263]1802e7d54f9aa51.png[/img] A: cos(5πt) B: cos(10πt) C: sin(5πt) D: sin(10πt)

（107）16= ( ) 2= ( )8= ( )8421BCD= ( )10

下列信号中，（ ）信号的频谱是连续的。 A: $x(t) = A\sin (\omega t + {\varphi _1}) + B\sin (3\omega t + {\varphi _2})$ B: $x(t) = 5\sin 30t + 3\sin \sqrt {50} t$ C: $x(t) = {e^{ - at}}\sin {\omega _0}t$