Answer:
The answer is "False".
Explanation:
ENIAC stands for Electronic Numerical Integrator and Computer, It was first developed in 1946. This computer use circuits and node that are interconnected with each other to show the result.
- ENIAC was a firstly, Reconfigurable vacuum tube, that controls the unit and it doesn't store information
- It takes a large room to set up the system, that's why it is not true.
Introductory program; just a static picture of a colored triangle.
Shows how to use GLUT.
Has minimal structure: only main() and a display callback.
Uses only the default viewing parameters (in fact, it never mentions viewing at all). This is an orthographic view volume with bounds of -1..1 in all three dimensions.
Draws only using glColor and glVertex within glBegin and glEnd in the display callback.
Uses only the GL_POLYGON drawing mode.
Illustrates glClear and glFlush.
triangle.cpp
// A simple introductory program; its main window contains a static picture
// of a triangle, whose three vertices are red, green and blue. The program
// illustrates viewing with default viewing parameters only.
#ifdef __APPLE_CC__
#include
#else
#include
#endif
// Clears the current window and draws a triangle.
void display() {
// Set every pixel in the frame buffer to the current clear color.
glClear(GL_COLOR_BUFFER_BIT);
// Drawing is done by specifying a sequence of vertices. The way these
// vertices are connected (or not connected) depends on the argument to
// glBegin. GL_POLYGON constructs a filled polygon.
glBegin(GL_POLYGON);
glColor3f(1, 0, 0); glVertex3f(-0.6, -0.75, 0.5);
glColor3f(0, 1, 0); glVertex3f(0.6, -0.75, 0);
glColor3f(0, 0, 1); glVertex3f(0, 0.75, 0);
glEnd();
// Flush drawing command buffer to make drawing happen as soon as possible.
glFlush();
}
// Initializes GLUT, the display mode, and main window; registers callbacks;
// enters the main event loop.
int main(int argc, char** argv) {
// Use a single buffered window in RGB mode (as opposed to a double-buffered
// window or color-index mode).
glutInit(&argc, argv);
glutInitDisplayMode(GLUT_SINGLE | GLUT_RGB);
// Position window at (80,80)-(480,380) and give it a title.
glutInitWindowPosition(80, 80);
glutInitWindowSize(400, 300);
glutCreateWindow("A Simple Triangle");
// Tell GLUT that whenever the main window needs to be repainted that it
// should call the function display().
glutDisplayFunc(display);
// Tell GLUT to start reading and processing events. This function
// never returns; the program only exits when the user closes the main
// window or kills the process.
glutMainLoop();
}
Answer:
A 1 MB digital file needs 0.23 seconds to transfer over a channel with bandwidth 10 MHz and SNR 10 dB.
Explanation:
We can calculate the channel capacity using Shannon's Capacity formula:
C = B + log₂ (1 + SNR)
Where C = Channel Capacity
B = Bandwidth of the Channel
SNR = Signal to Noise Ratio
We are given SNR in dB so we need to convert it into a ratio.
= 10log₁₀ (SNR)
10 = 10log₁₀ (SNR)
1 = log₁₀ (SNR)
SNR = 10¹
SNR = 10
So, using Shannon Channel Capacity formula:
C = 10 x 10⁶ log₂ (1 + 10)
C = 34.5 MHz
Total amount of time required to transmit a 1MB file:
1MB = 1 x 8 Mbytes = 8Mb
C = 34.5 MHz = 34.5 Mb/s
Time required = 8Mb/34.5Mb/s = 0.23 seconds
A 1 MB digital file needs 0.23 seconds to transfer over a channel with bandwidth 10 MHz and SNR 10 dB.
